Proposition (PageIndex{1})

Let ( riangle ABC) and ( riangle A'B'C') be two triangles in the neutral plane such that (AC = A'C') and (BC = B'C'). Then

(AB < A'B') if and only if (|measuredangle ACB| < |measuredangle A'C'B'|).

**Proof**Without loss of generality, we may assume that (A = A'), (C = C'), and (measuredangle ACB), (measuredangle ACB' ge 0). In this case we need to show that

(AB < AB' Leftrightarrow measuredangle ACB < measuredangle ACB'.)

Choose a point (X) so that

(measuredangle ACX = dfrac{1}{2} cdot (measuredangle ACB + measuredangle ACB').)

Note that

- ((CX)) bisects (angle BCB').
- ((CX)) is the perpendicular bisector of ([BB']).
- (A) and (B) lie on the same side of ((CX)) if and only if

(measuredangle ACB < measuredangle ACB').

From Exercise 5.2.1, (A) and (B) lie on the same side of ((CX)) if and only if (AB < AB'). Hence the result.

Theorem (PageIndex{1})

Let ( riangle ABC) be a triangle in the neutral plane. Then

(|measuredangle ABC| + |measuredangle BCA| + |measuredangle CAB| le pi.)

The following proof is due to Legendre [12], earler proofs were due to Saccheri [16] and Lambert [11].

**Proof**Set

(egin{array} {rclcrclcrcl} {a} & = & {BC,} & & {b} & = & {CA,} & & {c} & = & {AB,} {alpha} & = & {measuredangle CAB,} & & {eta} & = & {measuredangle ABC,} & & {gamma} & = & {measuredangle BCA.} end{array})

Without loss of generality, we may assume that (alpha, eta, gamma ge 0).

Fix a positive integer (n). Consider the points (A_0, A_1, ..., A_n) on the half-line ([BA)), such that (BA_i = i cdot c) for each (i). (In particular, (A_0 = B) and (A_1 = A).) Let us construct the points (C_1, C_2, ..., C_n), so that (measuredangle A_iA_{i-1}C_i = eta) and (A_{i-1} C_i = a) for each (i).

By SAS, we have constructed n congruent triangles

( riangle ABC = riangle A_1A_0C_1 cong riangle A_2A_1C_2 cong ... cong riangle A_nA_{n-1} C_n.)

Set (d = C_1C_2) and (delta = measuredangle C_2A_1C_1). Note that

[alpha + eta + delta = pi.]

By Proposition 11.2.1, we get that (Delta ge 0).

By construction

( riangle A_1C_1C_2 cong riangle A_2C_2C_3 cong ... cong riangle A_{n - 1} C_{n - 1} C_n.)

In particular, (C_i C_{i + 1} = d) for each (i).

By repeated application of the triangle inequality, we get that

(egin{array} {rcl} {n cdot c} & = & {A_0A_n le} {} & le & {A_0 C_1 + C_1 C_2 + cdots + C_{n - 1} C_n + C_n A_n =} {} & = & {a + (n - 1) cdot d + b.} end{array})

In particular,

(c le d + dfrac{1}{n} cdot (a + b - d).)

Since (n) is arbitrary positive integer, the latter implies (c le d). By Proposition (PageIndex{1}), it is equivalent to

(gamma le delta.)

From 11.3.1, the theorem follows.

Exercise (PageIndex{1})

Let (ABCD) be a quadrangle in the neutral plane. Suppose that the angles (DAB) and (ABC) are right. Show that (AB le CD).

**Hint**Set (a = AB, b = BC, c = CD), and (d = DA); we nned to show that (c ge a).

Mimic the proof of Theorem (PageIndex{1}) for the shown fence made from copies of quadrangle (ABCD).

## Math: How to Calculate the Angles in a Right Triangle

Every triangle has three sides, and three angles in the inside. These angles add up to 180° for every triangle, independent of the type of triangle. In a right triangle, one of the angles is exactly 90°. Such an angle is called a right angle.

To calculate the other angles we need the sine, cosine and tangent. In fact, the sine, cosine and tangent of an acute angle can be defined by the ratio between sides in a right triangle.

### Right Triangle

Just like every other triangle, a right triangle has three sides. One of them is the hypothenuse, which is the side opposite to the right angle. The other two sides are identified using one of the other two angles. The other angles are formed by the hypothenuse and one other side. This other side is called the adjacent side. Then, there is one side left which is called the opposite side. When you would look from the perspective of the other angle the adjacent and opposite side are flipped.

So if you look at the picture above, then the hypothenuse is denoted with h. When we look from the perspective of the angle alpha the adjacent side is called b, and the opposite side is called a. If we would look from the other non-right angle, then b is the opposite side and a would be the adjacent side.

### Sine, Cosine and Tangent

The sine, cosine and tangent can be defined using these notions of hypothenuse, adjacent side and opposite side. This only defines the sine, cosine and tangent of an acute angle. The sine, cosine and tangent are also defined for non-acute angles. To give the full definition, you will need the unit circle. However, in a right triangle all angles are non-acute, and we will not need this definition.

The sine of an acute angle is defined as the length of the opposite side divided by the length of the hypothenuse.

The cosine of an acute angle is defined as the length of the adjacent side divided by the length of the hypothenuse.

The tangent of an acute angle is defined as the length of the opposite side divided by the length of the adjacent side.

Or more clearly formulated:

### Calculating an Angle in a Right Triangle

The rules above allow us to do calculations with the angles, but to calculate them directly we need the inverse function. An inverse function f -1 of a function f has as input and output the opposite of the function f itself. So if f(x) = y then f -1 (y) = x.

So if we know sin(x) = y then x = sin -1 (y), cos(x) = y then x = cos -1 (y) and tan(x) = y then tan -1 (y) = x. Since these functions come up a lot they have special names. The inverse of the sine, cosine and tangent are the arcsine, arccosine and arctangent.

For more information on inverse functions and how to calculate them, I recommend my article about the inverse function.

### An Example of Calculating the Angles in a Triangle

In the triangle above we are going to calculate the angle theta. Let x = 3, y = 4. Then by the Pythagorean theorem we know that r = 5, since sqrt(3 2 + 4 2 ) = 5. Now we can calculate the angle theta in three different ways.

So theta = arcsin(3/5) = arccos(4/5) = arctan(3/4) = 36.87°. This allows us to calculate the other non-right angle as well, because this must be 180-90-36.87 = 53.13°. This is because the sum of all angles of a triangle always is 180°.

We can check this using the sine, cosine and tangent again. We call the angle alpha then:

Then alpha = arcsin(4/5) = arccos(3/5) = arctan(4/3) = 53.13. So this is indeed equal to the angle we calculated with the help of the other two angles.

We can also do it the other way around. When we know the angle and the length of one side, we can calculate the other sides. Let&aposs say we have a slide which is 4 meters long and goes down in an angle of 36°. Now we can calculate how much vertical and horizontal space this slide will take. We are basically in the same triangle again, but now we know theta is 36° and r = 4. Then to find the horizontal length x we can use the cosine. We get:

And therefore x = 4*cos(36) = 3.24 meters.

To calculate the height of the slide we can use the sine:

And therefore y = 4*sin(36) = 2.35 meters.

Now we can check whether tan(36) is indeed equal to 2.35/3.24. We find tan(36) = 0.73, and also 2.35/3.24 = 0.73. So indeed we did everything correctly.

### The Secant, Cosecant and Cotangent

The sine, cosine and tangent define three ratios between sides. There are however three more ratios we could calculate. If we divide the length of the hypothenuse by the length of the opposite is the cosecant. Dividing the hypothenuse by the adjacent side gives the secant and the adjacent side divided by the opposite side results in the cotangent.

This means that these quantities can be directly calculated from the sine, cosine and tangent. Namely:

The secant, cosecant and cotangent are used very rarely used, because with the same inputs we could also just use the sine, cosine and tangent. Therefore, a lot of people would not even know they exist.

### The Pythagorean Theorem

The Pythagorean Theorem is closely related to the sides of right triangles. It is very well known as a 2 + b 2 = c 2 . I wrote an article about the Pythagorean Theorem in which I went deep into this theorem and its proof.

## Angles in a Triangle Worksheets | Angle Sum Property, Exterior Angle Theorem

Angles in a triangle worksheets contain a multitude of pdfs to find the interior and exterior angles with measures offered as whole numbers and algebraic expressions. Learn to apply the angle sum property and the exterior angle theorem, solve for 'x' to determine the indicated interior and exterior angles. These printable exercises are customized for students of 6th grade through high school. Access some of these worksheets for free!

The angle sum property states that the interior angles of a triangle add up to 180°. Figure out if the given sets of angles form a triangle by adding them. Respond to each with 'Yes' or 'No'.

Subtract the sum of the two angles from 180° to find the measure of the indicated interior angle in each triangle.

Applying the exterior angle theorem, add the two opposite interior angles to find the unknown exterior angle of a triangle.

In these pdf worksheets, the measure of one of the interior angles of each triangle is presented as an algebraic expression. Set up an equation with the sum of the three angles, equating it with 180° and solve for 'x'.

Equate the sum of the two sides with the exterior angle depicted as an algebraic expression. Simplify the expression and find the value of 'x' in this stack of printable worksheets for grade 7 and grade 8.

Solve for 'x', substitute it in the expression(s) and find the measure of the indicated interior angle(s). The measures of two angles are offered as algebraic expressions in Part A and three angles in Part B.

Form an equation with the sum of the opposite angles with the exterior angle, simplify and find the value of 'x'. Plug it and compute the measure of the indicated angle in Part A and the measure of four angles in Part B.

Challenge high school students with the word format problems involving composite triangles containing right, isosceles and equilateral triangles. Determine the size of the indicated angles by applying the angle sum property and the exterior angle theorem.

## Download Go Math Grade 8 Chapter 11 Angle Relationships in Parallel Lines and Triangles Answer Key PDF

Enhance your skills by using the Go Math Grade 8 Answer Key Chapter 11 Angle Relationships in Parallel Lines and Triangles. Download Go Math Grade 8 Chapter 11 Answer Key and go through Questions and Answers on our website. Follow the below given Go Math Grade 8 Chapter 11 Angle Relationships in Parallel Lines and Triangles Answer Key topic wise links and start your preparation. Make use of the links and secure a good percentage in the exam.

**Lesson 1: Parallel Lines Cut by a Transversal**

**Lesson 2: Angle Theorems for Triangles**

**Lesson 3: Angle-Angle Similarity**

### Guided Practice – Parallel Lines Cut by a Transversal – Page No. 350

**Use the figure for Exercises 1–4.**

Question 1.

∠UVY and ____ are a pair of corresponding angles.

∠ _________

Explanation:

∠UVY and ∠ VWZ are a pair of corresponding angles.

When two lines are crossed by Transversal the angles in matching corners are called corresponding angles.

Question 2.

∠WVY and ∠VWT are _________ angles.

____________

Answer:

∠WVY and ∠VWT are alternate interior angles.

Alternate Interior Angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal.

Explanation:

∠WVY and ∠VWT are alternate interior angles.

Alternate Interior Angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal.

Question 3.

Find m∠SVW.

_________ °

Explanation:

∠SVW and ∠VWT are same sider interior angles. Therefore,

m∠SVW + m∠VWT = 180º

4xº +5xº = 180º

9x = 180º

x = 180/9

x = 20

m∠SVW = 4xº = (4.20)º = 80º

Question 4.

Find m∠VWT.

_________ °

Explanation:

∠SVW and ∠VWT are same sider interior angles. Therefore,

m∠SVW + m∠VWT = 180º

4xº +5xº = 180º

9x = 180º

x = 180/9

x = 20

m∠VWT = 5xº = (5.20)º = 100º

Question 5.

Vocabulary When two parallel lines are cut by a transversal, _______________ angles are supplementary.

____________

Answer:

If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent. If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles are supplementary.

**ESSENTIAL QUESTION CHECK-IN**

Question 6.

What can you conclude about the interior angles formed when two parallel lines are cut by a transversal?

Type below:

____________

Answer:

Alternate interior angles are congruent same-side interior angles are supplementary.

Explanation:

When two parallel lines are cut by a transversal, the interior angles will be the angles between the two parallel lines. Alternate interior angles will be on opposite sides of the transversal the measures of these angles are the same.

Same-side interior angles will be on the same side of the transversal the measures of these angles will be supplementary, adding up to 180 degrees.

### 11.1 Independent Practice – Parallel Lines Cut by a Transversal – Page No. 351

**Vocabulary Use the figure for Exercises 7–10.**

Question 7.

Name all pairs of corresponding angles.

Type below:

____________

Answer:

∠1 and ∠5

∠3 and ∠7

∠2 and ∠6

∠4 and ∠8

Explanation:

Corresponding angles are

∠1 and ∠5

∠3 and ∠7

∠2 and ∠6

∠4 and ∠8

Question 8.

Name both pairs of alternate exterior angles.

Type below:

____________

Explanation:

Alternate exterior angles

∠1 and ∠8

∠2 and ∠7

Question 9.

Name the relationship between ∠ 3 and ∠6.

Type below:

____________

Answer:

alternate interior angles

Explanation:

∠3 and ∠6 are alternate interior angles.

Alternate Interior Angles are a pair of angles on the inner side of each of those two lines but on opposite sides of the transversal.

Question 10.

Name the relationship between ∠4 and ∠6.

Type below:

____________

Answer:

same-side interior angles

Explanation:

∠4 and ∠6 are same-side interior angles.

**Find each angle measure.**

Question 11.

m∠AGE when m∠FHD = 30°

_________ °

Explanation:

∠AGE and ∠FHD are alternate exterior angles.

Therefore, m∠AGE = m∠FHD = 30°

m∠AGE = 30°

Question 12.

m∠AGH when m∠CHF = 150°

_________ °

Explanation:

∠AGH and ∠CHF are corresponding angles.

Therefore, m∠AGH = m∠CHF = 150°

m∠AGH = 150°

Question 13.

m∠CHF when m∠BGE = 110°

_________ °

Explanation:

∠CHF and ∠BGE are alternate exterior angles.

Therefore, m∠CHF = m∠BGE = 110°

m∠CHF = 110°

Question 14.

m∠CHG when m∠HGA = 120°

_________ °

Explanation:

∠CHF and ∠HGA are same-side interior angles.

m∠CHG + m∠HGA = 180°

m∠CHG + 120° = 180°

m∠CHG = 180 – 120 = 60

m∠CHG = 60º

Question 15.

m∠BGH

_________ °

Explanation:

∠BGH and ∠GHD are same-side interior angles.

So, ∠BGH + ∠GHD = 180º

3x + (2x + 50)º = 180º

5x = 180º – 50º = 130º

x = 130/5 = 26º

∠BGH = 3xº = 3 × 26º = 78º

∠GHD = (2x + 50) += (2 × 26 + 50) = 102º

Question 16.

m∠GHD

_________ °

Explanation:

∠BGH and ∠GHD are same-side interior angles.

So, ∠BGH + ∠GHD = 180º

3x + (2x + 50)º = 180º

5x = 180º – 50º = 130º

x = 130/5 = 26º

∠BGH = 3xº = 3 × 26º = 78º

∠GHD = (2x + 50) += (2 × 26 + 50) = 102º

Question 17.

The Cross Country Bike Trail follows a straight line where it crosses 350th and 360th Streets. The two streets are parallel to each other. What is the measure of the larger angle formed at the intersection of the bike trail and 360th Street? Explain.

_________ °

Answer:

The larger angle formed at the intersection of the bike trail and 360th Street is 132º

Explanation:

The larger angle formed at the intersection of the bike trail and 360th Street is the angle 5 in our schema. ∠5 and ∠3 are same-side interior angles. Therefore, m∠5 + m∠3 = 180º

m∠5 + 48º = 180º

m∠5 = 180º – 48º

m∠5 = 132º

Question 18.

Critical Thinking How many different angles would be formed by a transversal intersecting three parallel lines? How many different angle measures would there be?

_________ different angles

_________ different angle measures

Answer:

12 different angles

2 different angle measures

Explanation:

There are 12 different angles formed by a transversal intersecting three parallel lines.

There are 2 different angle measures:

m∠1 = m∠4 = m∠5 = m∠8 = m∠9 = m∠12

m∠2 = m∠3 = m∠6 = m∠7 = m∠10 = m∠11

### Parallel Lines Cut by a Transversal – Page No. 352

Question 19.

Communicate Mathematical Ideas In the diagram at the right, suppose m∠6 = 125°. Explain how to find the measures of each of the other seven numbered angles.

Type below:

____________

Answer:

m∠2 = m∠6 = 125º because ∠2 and ∠6 are corresponding angles.

m∠3 = m∠2 = 125º because ∠3 and ∠2 are vertical angles.

m∠7 = m∠3 = 125º because ∠7 and ∠3 are corresponding angles.

∠4 and ∠6 are same-side interior angles.

Therefore, m∠4 + m∠6 = 180º

m∠4 + 125º = 180º

m∠4 = 180º – 125º

m∠4 = 55º

m∠8 = m∠4 = 55º because ∠8 and ∠4 are corresponding angles.

m∠1 = m∠4 = 55º because ∠1 and ∠4 are vertical angles.

m∠5 = m∠1 = 55º because ∠5 and ∠1 are corresponding angles.

**FOCUS ON HIGHER ORDER THINKING**

Question 20.

Draw Conclusions In a diagram showing two parallel lines cut by a transversal, the measures of two same-side interior angles are both given as 3x°. Without writing and solving an equation, can you determine the measures of both angles? Explain. Then write and solve an equation to find the measures.

Answer:

m∠1 and m∠2 are same-side interior angles is 180º

Therefore, m∠1 + m∠2 = 180º

3x + 3x = 180º

6x = 180º

x = 180/6 = 30

m∠1 = m∠2 = 3x = 3(30) = 90º

Question 21.

Make a Conjecture Draw two parallel lines and a transversal. Choose one of the eight angles that are formed. How many of the other seven angles are congruent to the angle you selected? How many of the other seven angles are supplementary to your angle? Will your answer change if you select a different angle?

Type below:

____________

Answer:

We have to select ∠a form of eight angles that are formed. There are two other angles that are congruent to the angle ∠a. Two other angles are ∠e and ∠g.

There are no supplementary to ∠a.

If we select a different angle then the answer will also change.

Question 22.

Critique Reasoning In the diagram at the right, ∠2, ∠3, ∠5, and∠8 are all congruent, and∠1, ∠4, ∠6, and ∠7 are all congruent. Aiden says that this is enough information to conclude that the diagram shows two parallel lines cut by a transversal. Is he correct? Justify your answer.

____________

Answer:

This is not enough information to conclude that the diagram shows two parallel lines cut by a transversal. Because ∠2 and ∠3 are same-side interior angles. But ∠5 and ∠8 are not congruent with each other. And ∠6 and ∠7 are same-side interior angles. But ∠1 and ∠4 are not congruent with each other.

### Guided Practice – Angle Theorems for Triangles – Page No. 358

**Find each missing angle measure.**

Question 1.

m∠M = _________ °

Explanation:

From the Triangle Sum Theorem,

m∠L + m∠N + m∠M = 180º

78º + 31º + m∠M = 180º

109º + m∠M = 180º

m∠M = 180º – 109º

m∠M = 71º

Question 2.

m∠Q = _________ °

Explanation:

From the Triangle Sum Theorem,

m∠Q + m∠S + m∠R = 180º

m∠Q + 24º + 126º = 180º

m∠Q + 150º = 180º

m∠Q = 180º – 150º

m∠Q = 30º

**Use the Triangle Sum Theorem to find the measure of each angle in degrees.**

Question 3.

m∠T = _________ °

m∠V = _________ °

m∠U = _________ °

Answer:

m∠T = 88°

m∠V = 63°

m∠U = 29°

Explanation:

From the Triangle Sum Theorem,

m∠U + m∠T + m∠V = 180º

(2x + 5)º + (7x + 4)º + (5x + 3)º = 180º

2xº + 5º + 7xº + 4º + 5xº + 3º = 180º

14xº + 12º = 180º

14xº = 168º

x = 168/14 = 12

Substitute x value to find the angles

m∠U = (2x + 5)º = ((2 . 12) + 5)º = 29º

m∠U = 29º

m∠T = (7x + 4)º = ((7 . 12) + 4)º = 88º

m∠T = 88º

m∠V = (5x + 3)º = ((5 . 12) + 3)º = 63º

m∠V = 63º

Question 4.

m∠X = _________ °

m∠Y = _________ °

m∠Z = _________ °

Answer:

m∠X = 90°

m∠Y = 45 °

m∠Z = 45°

Explanation:

From the Triangle Sum Theorem,

m∠X + m∠Y + m∠Z = 180º

nº + (1/2 . n)º + (1/2 . n)º = 180º

2nº = 180º

n = 90

Substitute n values to find the angles

m∠X = nº = 90º

m∠X = 90º

m∠Y = (1/2 . n)º = (1/2 . 90)º = 45º

m∠Y = 45º

m∠Z = (1/2 . n)º = (1/2 . 90)º = 45º

m∠Z = 45º

**Use the Exterior Angle Theorem to find the measure of each angle in degrees.**

Question 5.

m∠C = _________ °

m∠D = _________ °

Explanation:

Given m∠C = 4y°, m∠D = (7y + 6)°, m∠E = 116°

By using exterior angle theorem,

∠DEC + ∠DEF = 180°

∠DEC + 116° = 180°

∠E = ∠DEC = 180° – 116° = 64°

The sum of the angles of a traingle = 180°

∠C +∠D + ∠E = 180°

4y° + (7y + 6)°+ 64° = 180°

11y° + 70° = 180°

11y° = 180° – 70° = 110°

y = 10

∠C = 4y° = 4. 10 = 40°

∠D = (7y + 6)° = ((7 . 10) + 6)° = (70 + 6)° = 76°

Question 6.

m∠L = _________ °

m∠M = _________ °

Explanation:

Given that m∠M = (5z – 3)°, m∠L = (18z + 3)°, m∠JKM = 161°

From the Exterior Angle Theorem,

m∠M + m∠L = m∠JKM

(5z – 3)° + (18z + 3)° = 161°

5z° – 3° + 18z° + 3° = 161°

23z° = 161°

z = 161/23 = 7

Substitute z values to find the angles

m∠M = (5z – 3)° = ((5 . 7) – 3)° = 32°

m∠L = (18z + 3)° = ((18 . 7) + 3)° = 129°

From the Triangle Sum Theorem,

m∠M + m∠L + m∠LKM = 180º

32º + 129º + m∠LKM = 180º

161º + m∠LKM = 180º

m∠LKM = 19º

**ESSENTIAL QUESTION CHECK-IN**

Question 7.

Describe the relationships among the measures of the angles of a triangle.

Type below:

______________

Answer:

The sum of all measures of the interior angles of a triangle is 180°. The measure of an exterior angle of a triangle is equal to the sum of its remote interior angles.

### 11.2 Independent Practice – Angle Theorems for Triangles – Page No. 359

**Find the measure of each angle.**

Question 8.

m∠E = _________ °

m∠F = _________ °

Explanation:

m∠E = x°, m∠F = x°, m∠D = 98°

From the Triangle Sum Theorem, sum of the angles of the traingle is 180°

m∠E + m∠D + m∠F = 180°

x + 98 + x = 180°

2x + 98 = 180°

2x = 82°

x = 41°

So, m∠E = 41°

m∠F = 41°

Question 9.

m∠T = _________ °

m∠V = _________ °

Explanation:

m∠W = 90°, m∠T = 2x°, m∠V = x°

From the Triangle Sum Theorem, sum of the angles of the traingle is 180°

m∠T + m∠V + m∠W = 180°

2x + x + 90 = 180°

3x = 90°

x = 30°

So, m∠T = 2x° = 2 . 30° = 60°

m∠V = x° = 30°

Question 10.

m∠G = _________ °

m∠H = _________ °

m∠J = _________ °

Answer:

m∠G = 75°

m∠H = 60°

m∠J = 45°

Explanation:

m∠G = 5x°, m∠H = 4x°, m∠J = 3x°

From the Triangle Sum Theorem, sum of the angles of the traingle is 180°

m∠G + m∠H + m∠J = 180°

5x + 4x + 3x = 180°

12x = 90°

x = 15°

So, m∠G = 5x° = 5 . 15° = 75°

m∠H = 4x° = 4. 15° = 60°

m∠J = 3x° = 3. 15° = 45°

Question 11.

m∠Q = _________ °

m∠P = _________ °

m∠QRP = _________ °

Answer:

m∠Q = 98°

m∠P = 55°

m∠QRP = 27°

Explanation:

Given that m∠Q = (3y + 5)°, m∠P = (2y – 7)°, m∠QRS = 153°

From the exterior angle Theorem,

∠QRS + ∠QRP = 180°

153° + ∠QRP = 180°

m∠R = m∠QRP = 180° – 153° = 27°

From the Triangle Sum Theorem, the sum of the angles of the triangle is 180°

m∠P + m∠Q + m∠R = 180°

(3y + 5)° + (2y – 7)°+ 27° = 180°

5y° + 25 = 180°

5y° = 155°

y = 31°

m∠Q = (3y + 5)° = ((3 . 31°) + 5)° = 98°

m∠P = (2y – 7)° = ((2. 31° – 7)° = 55°

m∠QRP = 27°

Question 12.

m∠ACB = _________ °

m∠DCE = _________ °

m∠BCD = _________ °

Answer:

m∠ACB = 44°

m∠DCE = 35°

m∠BCD = 101°

Explanation:

In traingle ABC, m∠A = 78°, m∠B = 58°, m∠ACB = ?°

From the Triangle Sum Theorem, the sum of the angles of the triangle is 180°

m∠A + m∠B + m∠ACB = 180°

78° + 58° + m∠ACB = 180°

m∠ACB = 180° – 136°

m∠ACB = 44°

In traingle CDE, m∠D = 85°, m∠E = 60°, m∠CDE = ?°

From the Triangle Sum Theorem, the sum of the angles of the triangle is 180°

m∠D + m∠E + m∠CDE = 180°

85° + 60° + m∠CDE = 180°

m∠CDE = 180° – 145°

m∠CDE = 35°

From the Exterior Angle Theorem,

m∠ACB + m∠CDE + m∠BCD = 180°

44° + 35° + m∠BCD = 180°

m∠BCD = 180° – 79°

m∠BCD = 101°

Question 13.

m∠K = _________ °

m∠L = _________ °

m∠KML = _________ °

m∠LMN = _________ °

Explanation:

m∠K = 2x°, m∠L = 3x°, m∠KML = x°

So, From the Triangle Sum Theorem, the sum of the angles of the triangle is 180°.

m∠K + m∠L + m∠KML = 180°

2x° + 3x° + x° = 180°

6x° = 180°

x= 30°

∠KML = x = 30°

∠L = 3x = 3 . 30° = 90°

∠K = 2x = 2 . 30° = 60°

From the Exterior Angle Theorem,

∠KML + ∠LMN = 180°

∠LMN = 180° – 30° = 150°

Question 14.

Multistep The second angle in a triangle is five times as large as the first. The third angle is two-thirds as large as the first. Find the angle measures.

The measure of the first angle: _________ °

The measure of the second angle: _________ °

The measure of the third angle: _________ °

Answer:

The measure of the first angle: 27°

The measure of the second angle: 135°

The measure of the third angle: 18°

Explanation:

Let us name the angles of a triangle as ∠1, ∠2, ∠3.

Consider ∠1 as x.

∠2 is 5 times as large as the first.

∠2 = 5x

Also, ∠3 = 2/3 . x

So, From the Triangle Sum Theorem, the sum of the angles of the triangle is 180°.

x+ 5x + (2/3 . x) = 180°

20x = 540°

x = 27°

So, ∠1 = x = 27°

∠2 = 5x = 5 . 27° = 135°

∠3 = 2/3 . x = 2/3 . 27° = 18°

The measure of the first angle: 27°

The measure of the second angle: 135°

The measure of the third angle: 18°

### Angle Theorems for Triangles – Page No. 360

Question 15.

Analyze Relationships Can a triangle have two obtuse angles? Explain.

___________

Answer:

No a triangle can NOT have two obtuse angles

Explanation:

The measure of an obtuse angle is greater than 90°. Two obtuse angles and the third angle would have a sum greater than 180°

**FOCUS ON HIGHER ORDER THINKING**

Question 16.

Critical Thinking Explain how you can use the Triangle Sum Theorem to find the measures of the angles of an equilateral triangle.

Type below:

___________

Answer:

All angles have the same measure in an equilateral triangle

Explanation:

Using the Triangle Sum Theorem,

∠x + ∠x + ∠x = 180°

3∠x = 180°

∠x = 60°

All angles have the same measure in an equilateral triangle

Question 17.

a. Draw Conclusions Find the sum of the measures of the angles in quadrilateral ABCD. (Hint: Draw diagonal (overline < AC >). How can you use the figures you have formed to find the sum?)

Sum: _________ °

Question 17.

b. Make a Conjecture Write a “Quadrilateral Sum Theorem.” Explain why you think it is true.

Type below:

___________

Answer:

The sum of the angle measures of a quadrilateral is 360°

Any quadrilateral can be divided into two triangles (180 + 180 = 360)

Question 18.

Communicate Mathematical Ideas Describe two ways that an exterior angle of a triangle is related to one or more of the interior angles.

Type below:

___________

Answer:

An exterior angle and it’s an adjacent interior angle equal 180°

An exterior angle equals the sum of the two remote interior angles.

### Guided Practice – Angle-Angle Similarity – Page No. 366

Question 1.

Explain whether the triangles are similar. Label the angle measures in the figure.

Type below:

___________

△ABC has angle measures _______and △DEF has angle measures______. Because _______in one triangle are congruent to ______in the other triangle, the triangles are_____.

Answer:

△ABC has angle measures 40°, 30°, and 109° and △DEF has angle measures 41°, 109°, and 30°. Because 2∠s in one triangle are congruent to in the other triangle, the triangles similar.

Question 2.

A flagpole casts a shadow 23.5 feet long. At the same time of day, Mrs. Gilbert, who is 5.5 feet tall, casts a shadow that is 7.5 feet long. How tall in feet is the flagpole? Round your answer to the nearest tenth.

_________ ft

Explanation:

In similar triangles, corresponding side lengths are proportional.

5.5/7.5 = h/23.5

h (7.5) = 129.25

h = 129.25/7.5

h = 17.23

Rounding to the nearest tenth

h = 17.2 feet

Question 3.

Two transversals intersect two parallel lines as shown. Explain whether △ABC and △DEC are similar.

∠BAC and∠EDC are ___________ since they are ___________.

∠ABC and∠DEC are ___________ since they are ___________.

By ________, △ABC and△DEC are ___________.

Type below:

___________

Answer:

∠BAC and∠EDC are congruent since they are alt. interior ∠s

∠ABC and∠DEC are congruent since they are alt. interior ∠s.

By AA similarity, △ABC and△DEC are similar.

**ESSENTIAL QUESTION CHECK-IN**

Question 4.

How can you determine when two triangles are similar?

Type below:

___________

Answer:

If 2 angles of one triangle are congruent to 2 angles of another triangle, the triangles are similar by the Angle-Angle Similarity Postulate

### 11.3 Independent Practice – Angle-Angle Similarity – Page No. 367

**Use the diagrams for Exercises 5–7.**

Question 5.

Find the missing angle measures in the triangles.

Type below:

___________

Answer:

m∠B = 42°

m∠F = 69°

m∠H = 64°

m∠K = 53°

Explanation:

Using the Triangle Sum Theorem,

m∠A + m∠B + m∠C = 180°

85° + m∠B + 53° = 180°

138° + m∠B = 180°

m∠B = 180° – 138°

m∠B = 42°

Using the Triangle Sum Theorem,

m∠D + m∠E + m∠F = 180°

We substitute the given angle measures and we solve for m∠F

64° + 47° + m∠F = 180°

111° + m∠F = 180°

m∠F = 180° – 111°

m∠F = 69°

Using the Triangle Sum Theorem,

m∠G + m∠H + m∠J = 180°

We substitute the given angle measures and we solve for m∠H

47° + m∠H + 69° = 180°

116° + m∠H = 180°

m∠H = 180° – 116°

m∠H = 64°

Using the Triangle Sum Theorem,

m∠J + m∠K + m∠L = 180°

We substitute the given angle measures and we solve for m∠K

85° + m∠K + 42° = 180°

127° + m∠K = 180°

m∠K = 180° – 127°

m∠K = 53°

Question 6.

Which triangles are similar?

Type below:

___________

Answer:

△ABC and △JKL are similar because their corresponding angles are congruent. Also, △DEF and △GHJ are similar because their corresponding is congruent.

Question 7.

Analyze Relationships Determine which angles are congruent to the angles in △ABC.

∠A ≅ ∠ ________

∠B ≅ ∠ ________

∠C ≅ ∠ ________

Explanation:

△JKL has angle measures that are the same as those is △ABC

∠A ≅ ∠ J

∠B ≅ ∠ L

∠C ≅ ∠ K

Therefore, they are congruent.

Question 8.

Multistep A tree casts a shadow that is 20 feet long. Frank is 6 feet tall,and while standing next to the tree he casts a shadow that is 4 feet long.

a. How tall is the tree?

h = ________ ft

Explanation:

In similar triangles, corresponding side lengths are proportional.

20/4 = h/6

5 = h/6

h = 30

The tree is 30 feet tall.

Question 8.

b. How much taller is the tree than Frank?

________ ft

Explanation:

30 – 6 = 24

The tree is 24 feet taller than Frank.

Question 9.

Represent Real-World Problems Sheila is climbing on a ladder that is attached against the side of a jungle gym wall. She is 5 feet off the ground and 3 feet from the base of the ladder, which is 15 feet from the wall. Draw a diagram to help you solve the problem. How high up the wall is the top of the ladder?

________ ft

3/15 = 5/h

15 ×3 = 3h

75 = 3h

h = 75/3 = 25

Question 10.

Justify Reasoning Are two equilateral triangles always similar? Explain.

______________

Answer:

yes two equilateral triangles are always similar.

Each angle of an equilateral triangle is 60°. Since both triangles are equilateral then they are similar.

### Angle-Angle Similarity – Page No. 368

Question 11.

Critique Reasoning Ryan calculated the missing measure in the diagram shown. What was his mistake?

(frac<3.4><6.5>=frac

19.5 × (frac<3.4><6.5>=frac

(frac<66.3><6.5>) = h

10.2cm = h

Type below:

___________

Answer:

In the first line, Ryan did not take the sum of 6.5 and 19.5 to get the denominator on the right.

The denominator on the right should be 26 instead of 19.5

the correct value for h

3.4/6.5 = h/26

h = (3.4/6.5) × 26

h = 13.6cm

**FOCUS ON HIGHER ORDER THINKING**

Question 12.

Communicate Mathematical Ideas For a pair of triangular earrings, how can you tell if they are similar? How can you tell if they are congruent?

Type below:

___________

Answer:

The earrings are similar if two angle measures of one are equal to two angle measures of the other.

The earrings are congruent if they are similar and if the side lengths of one are equal to the side lengths of the other.

Question 13.

Critical Thinking When does it make sense to use similar triangles to measure the height and length of objects in real life?

Type below:

___________

Answer:

If the item is too tall or the distance is too long to measure directly, similar triangles can help with measuring.

Question 14.

Justify Reasoning Two right triangles on a coordinate plane are similar but not congruent. Each of the legs of both triangles are extended by 1 unit, creating two new right triangles. Are the resulting triangles similar? Explain using an example.

___________

Answer:

Two triangles are similar if their corresponding angles are congruent and the lengths of their corresponding sides are proportional. If each of the legs of both triangles is extended by 1 unit, the ratio between proportional sides does not change. Therefore, the resulting triangles are similar.

### Ready to Go On? – Model Quiz – Page No. 369

**11.1 Parallel Lines Cut by a Transversal**

**In the figure, line p || line q. Find the measure of each angle if m∠8 = 115°.**

Explanation:

According to the exterior angle theorem,

m∠7 + m∠8 = 180°

m∠7 + 115° = 180°

m∠7 = 180° – 115°

m∠7 = 65°

Explanation:

From the given figure, Line P is parallel to line Q. So, the angles given in line P is equal to the angles in line Q. They are corresponding angles.

So, m∠8 is parallel is m∠6 or m∠8 = m∠6 = 115°

Explanation:

∠1 and ∠6 are alternative exterior angles.

So, m∠1 = m∠6 = 115°

**11.2 Angle Theorems for Triangles**

**Find the measure of each angle.**

Explanation:

m∠A + m∠B + m∠C = 180°

4y° + (3y + 22)° + 74° = 180°

7y = 180 – 96 = 84

y = 12°

m∠A = 4y° = 4 (12°) = 48°

m∠B = (3y + 22)° = (3(12°) + 22)° = 58°

Explanation:

m∠A + m∠B + m∠C = 180°

4y° + (3y + 22)° + 74° = 180°

7y = 180 – 96 = 84

y = 12°

m∠A = 4y° = 4 (12°) = 48°

m∠B = (3y + 22)° = (3(12°) + 22)° = 58°

Question 6.

m∠BCA = _________ °

Explanation:

m∠BCD + m∠BCA = 180°

106° + m∠BCA = 180°

m∠BCA = 180° – 106°

m∠BCA = 74°

So, m∠BCA = 74°

**11.3 Angle-Angle Similarity**

**Triangle FEG is similar to triangle IHJ. Find the missing values.**

Explanation:

In similar triangles, corresponding side lengths are proportional.

HJ/EG = IJ/FG

(x + 12)/42 = 40/60

(x + 12)/42 = 4/6

6x = 96

x = 16

Explanation:

In similar triangles, corresponding side lengths are congruent.

m∠HJI = m∠EGF

(5y + 7)° = 52°

5y° + 7° = 52°

5y° = 45°

y = 9

Explanation:

Using the Triangle Sum Theorem,

m∠E + m∠F + m∠G = 180°

We substitute the given angle measures and we solve for m∠E

m∠E + 36° + 52° = 180°

m∠E + 88° = 180°

m∠E = 92°

In similar angles, corresponding side lengths are congruent

m∠H = m∠E

m∠H = 92°

**ESSENTIAL QUESTION**

Question 10.

How can you use similar triangles to solve real-world problems?

Type below:

____________

Answer:

we know that if two triangles are similar, then their corresponding angles are congruent and the lengths of their corresponding sides are proportional. We can use this to determine values that we cannot measure directly. For example, we can calculate the length of the tree if we measure its shadow and our shadow on a sunny day.

### Selected Response – Mixed Review – Page No. 370

**Use the figure for Exercises 1 and 2.**

Question 1.

Which angle pair is a pair of alternate exterior angles?

Options:

A. ∠5 and ∠6

B. ∠6 and∠7

C. ∠5 and ∠4

D. ∠5 and ∠2

Explanation:

∠5 and ∠4 are alternate exterior angles

Question 2.

Which of the following angles is not congruent to ∠3?

Options:

A. ∠1

B. ∠2

C. ∠6

D. ∠8

Explanation:

∠2 and ∠3 are same-side interior angles. They are not congruent instead their sum is equal to 180°

Question 3.

The measures, in degrees, of the three angles of a triangle are given by 2x + 1, 3x – 3, and 9x. What is the measure of the smallest angle?

Options:

A. 13°

B. 27°

C. 36°

D. 117°

Explanation:

From the Triangle Sum Theorem, the sum of the angles of the triangle is 180°

m∠1 + m∠2 + m∠3 = 180°

(2x + 1)° + (3x – 3)° + (9x)° = 180°

2x° + 1° + 3x° – 3° + 9x° = 180°

14x° – 2° = 180°

14x° = 178°

x = 13

Substitute the value of x to find the m∠1, m∠2, and m∠3

m∠1 = (2x + 1)° = (2(13) + 1)° = 27°

m∠2 = (3x – 3)° = (3(13) – 3)° = 36°

m∠3 = (9x)° = (9(13))° = 117°

The smallest angle is 27°

Question 4.

Which is a possible measure of ∠DCA in the triangle below?

Options:

A. 36°

B. 38°

C. 40°

D 70°

Explanation:

Using the Exterior Angle Theorem

m∠A + m∠B = m∠DCA

m∠A + 40° = m∠DCA

m∠DCA will be greater than 40°. The only suitable option is D, 70°.

Question 5.

Kaylee wrote in her dinosaur report that the Jurassic period was 1.75 × 10 8 years ago. What is this number written in standard form?

Options:

A. 1,750,000

B. 17,500,000

C. 175,000,000

D. 17,500,000,000

Explanation:

1.75 × 10 8 standard form

Move the decimal point to 8 right places.

175,000,000

Question 6.

Given that y is proportional to x, what linear equation can you write if y is 16 when x is 20?

Options:

A. y = 20x

B. y = (frac<5><4>) x

C. y = (frac<4><5>)x

D. y = 0.6x

Explanation:

Y=4/5x

16=4/5(20)

4/5×20/1=80/5

80/5=16

Question 7.

Two transversals intersect two parallel lines as shown.

a. What is the value of x?

x = ________

Explanation:

m ∠ J K L = m ∠ L N M

6x + 1 = 25

6x = 24

x = 4

Question 7.

b. What is the measure of ∠LMN?

_________°

Explanation:

m∠LMN = 3x + 11 = 3(4) + 11 = 12 + 11 = 23

Question 7.

c. What is the measure of ∠KLM?

∠KLM = _________°

Explanation:

∠KLM exterior angle of the triangle LMN

m∠KLM = m∠LNM + m∠LMN

= 25 + 23 = 48

Question 7.

d. Which two triangles are similar? How do you know?

Type below:

_____________

Answer:

triangle JKL = triangle LNM

triangle KJL = triangle LMN

Explanation:

triangle JLK and triangle LNM are similar.

triangle JKL = triangle LNM

triangle KJL = triangle LMN

### Summary:

The solutions provided in the Go Math Grade 8 Answer Key Chapter 11 Angle Relationships in Parallel Lines and Triangles are made by the professionals. Practice all the math questions available on the 8th Grade Text Book and learn how to solve the questions in a simple way. Hope the information provided in this article is beneficial for all the students of grade 8. Keep in touch with our website to get the pdfs of all the Go Math Grade 8 Answer Key Chapterwise.

## Classifying triangles

Triangles are often classified by their angles and sides, as shown in the tables below.

### By angles:

Type | Angles | Figure |
---|---|---|

Acute | all interior angles < 90° | |

Obtuse | 1 interior angle > 90° | |

Right | 1 angle = 90° | |

Equiangular | each interior angle = 60° |

### By sides:

Type | Sides | Figure |
---|---|---|

Scalene | no 2 sides are congruent | |

Isosceles | 2 congruent sides | |

Equilateral | all sides are congruent |

## Fibonacci Fractals

Now we will explore the formation of spirals in more detail, and discover some more interesting and useful facts about Fibonacci Numbers. We will be examining the concept of *periodicity*, which we'll also learn about in the Mandelbrot Set, but which is also important in a large number of natural systems.

Perhaps the simplest, most elegant example of a spiral in nature is a seashell. The organism that creates the shell just repeats a simple process again and again to form the spiral shell. It keeps adding wedges to its shell in a very simple fashion: Each wedge is rotated by the same angle, and each wedge is the same proportion larger than the one before it. This is all that is required to make a logarithmic spiral. Notice that you could choose any angle or proportion, as long as you keep repeating the same values at each iteration.

We're going to explore what happens when we change the angle that generates the spiral. Play with the Angle slider in the Spiralizer applet below:

At very low angles, the applet creates a simple spiral. But as you increase the angle, all sorts of interesting patterns emerge. It can become difficult to determine the order of the dots. Click on "Connect Dots" to make the connections easier to see.

If you set the angle to 180 degrees, the point will rotate to the other side, and then back again at the next iteration, and so on, oscillating with a period of 2. If you set the angle to be 90 degrees, The dots will grow in a square pattern, that is, with a period of 4. The periodicity can be determined by dividing the angle of a full circle, 360 degrees, by the rotation angle. For example, 360 / 90 = 4. The progression of the points is just 0, 90, 180, 270 degrees. Then it returns to 360, which is the same as 0 degrees, and the pattern begins again.

There is another angle that will generate period-4 patterns: 270 degrees. This is 360 degrees - 90 degrees. We can see why this works by examining the progression of the angles. Start with iteration 0 at at 0 degrees. The first iteration takes the point 3/4 of the way around the circle or 270 degrees. Then the second iteration adds 270 degrees, which is 540 degrees. This is the same as 360 + 180, so it's halfway around the circle. The next iteration takes the point to 810 degrees, which is the same angle as 720 + 90, or 2 and 1/4 times around the circle.

In general, the periodicity **P = 360 / A**, where is the angle around the circle. If we want to find a specific periodicity, we can rearrange the equation to solve for A = 360 / P .

One interesting thing to observe is that when the angle is set exactly to a value that repeats perfectly, such as 90 degrees, the points line up in straight lines. But if you adjust the angle just a tiny bit larger or smaller than one of these perfect values, then the dots start to twist into spirals. You can nudge the angle 0.1 degrees at a time with the arrow keys to see the effect of *perturbing* the angle just slightly.

Questions:

Find an angle that generates a perfect triangle pattern (period 3): [ ]

Find another angle that generates a perfect triangle pattern: [ ]

Find an angle that generates a perfect pentagon pattern (period 5): [ ]

Find another angle that generates a perfect pentagon pattern: [ ]

Find an angle that generates a perfect hexagon pattern (period 6): [ ]

Find another angle that generates a perfect hexagon pattern: [ ]

There are many other higher-order periodicities that emerge in between the simple angles that create triangles, square, pentagons and hexagons, and we will explore the more complex details in the next section, when we relate this behavior to the periodicities of the Mandelbrot Set.

## Fibonacci Packing

A sunflower pattern can be created by a simple repetitive process similar to how the spiralizer forms its patterns. Imagine the flower operating like the Spiralizer. It creates a seed at the origin, and then it rotates by a certain angle and creates another seed. Then it rotates again by the same angle and forms a third seed. It keeps rotating by the same angle and adding seeds, which all keep growing in scale and distance from the center.

How does the angle affect the outcome of the pattern of seeds? As we saw with the spiralizer, when you set the angle to be a simple fraction of the whole way around the circle, the dots (or seeds) line up in radial arms. This is a simple pattern, but it is NOT the most efficient way to pack a lot of seeds into a given area. You can see the percentage of space filled by dots at the bottom of the Spiralizer. For instance, when the angle of rotation is 90 degrees, the dots fall in a period-4 pattern, and the percentage of the space filled by dots is 11.436%. This is a relatively small proportion, and is a poor use of space.

When the angle of rotation is not simple like 1/2, 1/3, or 1/4, but instead is an *irrational* fraction of the circle, then the angle never quite repeats itself, and much more complex patterns are possible. Sometimes these arrangements alloiw a much more efficient packing of seeds into a given area.

When the angle of rotation is set to be the 360/**&phi** that is, the fraction of the circle correspinding to the Golden Ratio, then the seeds pack in the most efficient way possible. Try it with the Spiralizer above! What is the "Golden Angle" for this optimal packing? **360 / 1.61803399** is approximately equal to **222.5** degrees.

Set the angle in the Spiralizer to 222.5 (you might need to use the arrow keys to set it precisely), turn the drawing speed down about half way, and click "Connect Dots" to illustrate how a sunflower arranges its seeds.

One amazing thing to observe is that in systems that use this packing system (many flowers, pinecones, strawberries, pineapples, artichokes, etc) you can often find the Fibonacci Sequence. This is illustrated in the picture of the sunflower above. The pattern forms intersecting spirals where the number of seeds in the counter-clockwise spiral is part of the Fibonacci Sequence, and the number of seeds in the clockwise spiral is the next highest Fibonacci Number. This is often difficult to count, and sometimes it is not exactly correct, but in general the *ratio* of the seeds counted in one direction to the number of seeds in the other direction is close to the Golden Ratio **&phi**.

Questions:

What percentage of the space is filled with dots for the Golden Angle: [ ] (Make sure "Connect Dots" is off.)

What percentage of the space is filled for the period-8 arrangement of dots? [ ]

What angle is the LEAST efficient for packing dots (i.e fills up the lowest proportion of space)? [ ]

In the sunflower image above, how many seeds are in the counter-clockwise spiral highlighted in blue? [ ]

## Answers and Replies

I think M is any point on the line connecting K and L. Using M, we can construct another line segment JM that connects J and M.

so the line segment JM has a length of JM and the KM and LM line segments have lengths of KM and LM respectively.

The question wants you answer what angle must J be such that:

JM * JM = KM * LM for any point M on the KL line segment

where the asterisk is simply real number multiplication and JM, KM, and LM are length measures.

I think M is any point on the line connecting K and L. Using M, we can construct another line segment JM that connects J and M.

so the line segment JM has a length of JM and the KM and LM line segments have lengths of KM and LM respectively.

The question wants you answer what angle must J be such that:

JM * JM = KM * LM for any point M on the KL line segment

where the asterisk is simply real number multiplication and JM, KM, and LM are length measures.

This makes it a lot easier. Only one thing is it really does not specify M is the mid point. But it won't work if it doesn't.

This is my work, let me know whether I am right. It's not hard after I understand the question.

I don't get use to angle J or angle K. I am more used to angle KJL and angle JKL. That would be a lot clearer.

If M is not the mid point, it's going to be harder to find.

Seems like if M is not the mid point, you can make the equation work with any angle less than 180deg. I don't know how to proof it, but there should be some length of the sides of the triangle to make the equation fit. So all the answers are correct in the problem. But I don't know how to proof it.

I tried to use cosine law, but there's too many variables. Anyone know of a way to proof this if M is not a mid point?

If the question means that for **any** point M between K and L it holds ##JM^2=KMcdot LM## then you have solve it because since it holds for any point, it holds for the mid point as well and from that you proved that J is 90. So J is 90 the only correct answer.

However I am not sure that the question means that. I think it means that there exists a point M (not necessarily the mid point ) **such that** ##JM^2=KMcdot LM##. Then it is totally different. Then 90 degrees is one possible answers but there might others as well and your answer is incomplete.

BTW in the last line of your work you meant to write that ##KJL=eta+gamma ## right?

Seems like if M is not the mid point, you can make the equation work with any angle less than 180deg. I don't know how to proof it, but there should be some length of the sides of the triangle to make the equation fit. So all the answers are correct in the problem. But I don't know how to proof it.

I tried to use cosine law, but there's too many variables. Anyone know of a way to proof this if M is not a mid point?

If it were my test question, I'd check all of them as I agree the angle can be any from 0 to 180 if the point M is anywhere along the segment KL. If you have access to Mathematica, perhaps you can study the code below with KL=1 and as the parameters "t" (angle of MJ on the circle -- the red line) and "a" (length of MJ) are varied we always have red^2=(green)(blue) and since we can scale it for any length KL then we can adjust the triangle with the angle at J to be 0 to 180. Then perhaps formulate it into a proof.

Well, I’m going to add my two pennies worth…

The question is incomplete. It is easy to construct a counter-example to demonstrate that - unless M is the midpoint - m∠J can have a wide range of values. This means m∠J can not be limited to one of the values in the given list.

Counter example:

Point K is at (0,0). Point L is at (10,0). Point M is (say) at (1, 0).

This gives KM = 1 and LM = 9 so that KM•LM = 1•9 = 9.

JM² = KM•LM = 1•9 = 9 which gives JM = 3.

Point J can therefore be anywhere on a circle centred on M and radius 3. A range of values for m∠J is therefore possible.

However if M is the midpoint of KL, then let r = KM = LM.

JM² = KM•LM = r² which gives JM = r.

This mean KL is the diameter of a circle centred on M and radius r, and J lies on this circle. ∠J is the angle subtended by a diameter and is therefore 90º (Thales’ theorem):

https://www.cut-the-knot.org/Outline/Geometry/AngleOnDiameter.jpg

So it seems most likely that the question neglected to state that M is the midpoint of KL.

## What are the Angles of a Triangle?

**The angle of a triangle is the space formed between two side lengths of a triangle.** A triangle contains interior angles and exterior angles. **Interior angles** are three angles found inside a triangle. **Exterior angles** are formed when the sides of a triangle are extended to infinity.

Therefore, exterior angles are formed outside a triangle between one side of a triangle and the extended side. Each exterior angle is adjacent to an interior angle. Adjacent angles are angles with a common vertex and side.

The figure below shows the **angle of a triangle**. The interior angles are a, b and c, while exterior angles are d, e, and f.

## Triangles Side and Angles

Triangles are one of the most fundamental geometric shapes and have a variety of often studied properties including:

Rule 1: Interior Angles sum up to $ 180^0 $ Rule 2: Sides of Triangle -- Triangle Inequality Theorem : This theorem states that the sum of the lengths of any 2 sides of a triangle must be greater than the third side. ) Rule 3: Relationship between measurement of the sides and angles in a triangle: The largest interior angle and side are opposite each other. The same rule applies to the smallest sized angle and side, and the middle sized angle and side. Rule 4 Remote Extior Angles -- This Theorem states that the measure of a an exterior angle $ angle A$ equals the sum of the remote interior angles' measurements. more)#### What's the difference between interior and exterior angles of a triangle?

This question is answered by the picture below. You create an exterior angle by extending any side of the triangle.

### Interior Angles of a Triangle Rule

This may be one the most well known mathematical rules-*The sum of all 3 interior angles in a triangle is $180^ $.* As you can see from the picture below, if you add up all of the angles in a triangle the sum must equal $180^

To explore the truth of this rule, try Math Warehouse's interactive triangle, which allows you to drag around the different sides of a triangle and explore the relationship between the angles and sides. No matter how you position the three sides of the triangle, the total degrees of all interior angles (the three angles inside the triangle) is always 180°.

This property of a triangle's interior angles is simply a specific example of the general rule for any polygon's interior angles.

### Interior Angles Interactive Demonstration

### Practice Problems (interior angles rule)

##### Problem 1

What is m$angle$LNM in the triangle below?

m$ angle $ LNM +m$ angle $ LMN +m$ angle $ MLN =180°

m$ angle $ LNM +34° + 29° =180°

m$ angle $ LNM +63° =180°

m$ angle $ LNM = 180° - 63° = **117°**

##### Problem 2

A triangle's interior angles are $ angle $ HOP, $ angle $ HPO and $ angle $ PHO. $ angle $ HOP is 64° and m$ angle $ HPO is 26°.

What is m$ angle $ PHO?

m$ angle $ PHO = 180° - 26° -64° = 90°

### Relationship --Side and Angle Angle Measurements

- the largest interior angle is
**opposite**the largest side - the smallest interior angle is
**opposite**the smallest side - the middle-sized interior angle is
**opposite**the middle-sized side

To explore the truth of the statements you can use Math Warehouse's interactive triangle, which allows you to drag around the different sides of a triangle and explore the relationships betwen the measures of angles and sides. No matter how you position the three sides of the triangle, you will find that the statements in the paragraph above hold true.

(All right, the isosceles and equilateral triangle are exceptions due to the fact that they don't have a single smallest side or, in the case of the equilateral triangle, even a largest side. Nonetheless, the principle stated above still holds true. !)

## 11.3: Three angles of triangle

**Input 3 triangle side lengths (A, B and C), then click "ENTER". This calculator will determine whether those 3 sides will form an equilateral, isoceles, acute, right or obtuse triangle or no triangle at all.**

** **

**Without Using The Calculator When given 3 triangle sides, to determine if the triangle is acute, right or obtuse:**

** **

**2) Sum the squares of the 2 shortest sides.**

3) Compare this sum to the square of the 3rd side.

if sum > 3rd side² Acute Triangle

if sum = 3rd side² Right Triangle

sum of the squares of the short sides = 25 + 36 = 61

3, 4 and 5 Squaring each side = 9, 16 and 25

sum of the squares of the short sides = 9 + 16 = 25

3, 7 and 9 Squaring each side = 9, 49 and 81

sum of the squares of the short sides = 9 + 49 = 58

58 For determining if the 3 sides can even form a triangle, the triangle inequality theorem states that the longest side must be shorter than the sum of the other 2 sides.