EXTERIOR ANGLE THEOREM OF TRIANGLE

Problem 1 :

Solve for x.

Solution :

Step 1 :

Write the Exterior Angle Theorem as it applies to this triangle.

m ∠ A + m ∠ B = m ∠ ACD

Step 2 :

Substitute the given angle measures.

21^º + 34^º = x^º

Step 3 :

55^º = x^º

So, the value of x is 55^º.

Problem 2 :

Solve for x and find m ∠A.

Solution :

Step 1 :

Write the Exterior Angle Theorem as it applies to this triangle.

m ∠A + m ∠B = m ∠C

Step 2 :

Substitute the given angle measures.

(2x + 3)^º + 51^º = 100^º

Step 3 :

Solve the equation for x.

(2x + 3)^º + 51^º = 100^º

2x + 3 + 51 = 100

2x + 54 = 100

Subtract 54 from both sides.

2x + 54 – 54 = 100 – 54

2x = 46

Divide both sides by 2.

2x/2 = 46/2

x = 23

Step 4 :

Use the value of x to find m ∠A.

m ∠A = 2x + 3

= 2(23) + 2

= 46 + 2

m ∠ A= 48

So, the value of m ∠A is 48^º.

Problem 3 :

Solve for x and find m ∠B.

Solution :

Step 1 :

Write the Exterior Angle Theorem as it applies to this triangle.

m ∠A + m ∠B = m ∠C

Step 2 :

Substitute the given angle measures.

60^º + 2x^º = 94^º

Subtract 60^º from both sides.

60^º - 60^º + 2x^º = 94^º - 60^º

2x^º = 34

Divide both sides by 2.

2x^º/2 = 34/2

x^º = 17

Step 3 :

Use the value of x to find m ∠ B.

m ∠ B = 2xº

= 2(17)

= 34

So, the value of m ∠B is 34^º.

Problem 4 :

Solve for x.

Solution :

Step 1 :

m ∠DCA = m ∠CAB + m ∠ABC

Step 2 :

Substitute the given angle measures.

6x - 7 = 2x + 103 - x

6x - 7 = x + 103

6x - x = 103 + 7

5x = 110

x = 110/5

x = 22

Problem 5 :

Solve for x.

Solution :

Step 1 :

Write the Exterior Angle Theorem as it applies to this triangle.

In a triangle ∠ABC

m ∠A + m ∠B + m ∠C = 180^º

Step 2 :

Substitute the given angle measures.

50^º + 62^º + m ∠C = 180^º

112 + m ∠C = 180^º

Subtract 112 from both sides.

m ∠C = 180 - 112

x = 68

Step 3 :

In a triangle ∠DEF

m ∠D + m ∠E + m ∠F = 180^º

Step 4 :

Substitute the given angle measures.

53^º + m ∠E + 80^º = 180^º

133 + m ∠E = 180^º

m ∠E = 180 - 133

m ∠E = 47

Step 5 :

∠EOC = 180 – (∠ABC + ∠DEF)

= 180 - 115

= 65

So, the value of x is 65.

Problem 6 :

Solve for x, find m ∠ A.

Solution :

Step 1 :

Write the Exterior Angle Theorem as it applies to this triangle.

m ∠A + m ∠B = m ∠C

Step 2 :

Substitute the given angle measures.

x^º + 90^º = 122^º

Step 3 :

x + 90 = 122

Subtract 90 from both sides.

x + 90 – 90 = 122 - 90

x = 32

Step 4 :

Use the value of x to find m ∠A.

m ∠A = xº

= 32

So, the value of m ∠ A is 32.

Problem 7 :

Solve for x.

Solution :

In triangle ABC,

∠ABC = ∠ACB

∠BAC = 56^º

∠CDA = x^º

∠ABC + ∠ACB + ∠BAC = 180

2∠ABC + 56 = 180

2∠ABC = 180 - 56

2∠ABC = 124

∠ABC = 124/2

∠ABC = 62

∠ACD = 56 + 62 ==> 118

∠CAD = ∠CDA = x

x + x + 118 = 180

2x + 118 = 180

2x = 62

x = 31

Determine the unknown angle measures in the figure.

Problem 8 :

Describe and correct the error in finding m∠1.

Solution :

Sum of interior angles of triangle = 180 degree

115 + 39 + m∠1 = 180

154 + m∠1 = 180

m∠1 = 180 - 154

m∠1 = 26

Problem 9 :

Describe and correct the error in finding m∠1.

Solution :

Exterior angle = sum of remote interior angles

m∠1 = 80 + 50

m∠1 = 130

So, the measure of exterior angle is 130 degree.

Problem 10 :

The measure of one of the base angles of an isosceles triangle is 42°. The measure of an exterior angle at the vertex of the triangle is

1) 42°        2) 84°         3) 96°       4) 138°

Solution :

Since it is isosceles triangle, the same side will make equal angle measures. Let 42 degree the equal angle measure, then the third angle measure will be x

42 + 42 + x = 180

84 + x = 180

x = 180 - 84

= 96

Option 3) is correct.