Finding Missing Angles in Triangles

Write and solve an equation to find the value of x for each of the triangle

Problem 1 :

Solution :

Sum of the interior angles of the triangle = 180

2x + 7 + 63 + 30 = 180

2x + 100 = 180

Subtracting 100 on both sides.

2x = 180 - 100

2x = 80

Divide by 2 on both sides.

x = 40

Problem 2 :

Solution :

Sum of interior angles = 180

115 + x + x = 180

2x + 115 = 180

Subtract 115 on both sides.

2x = 180 - 115

2x = 65

Divide it by 2.

x = 65/2

Problem 3 :

Solution:

60 + 60 + 4x - 80 = 180

120 + 4x - 80 = 180

40 + 4x = 180

Subtract 40 on both sides

4x = 180 - 40

4x = 140

Divide by 4 on both sides.

x = 140/4

x = 35

Problem 4 :

Solution :

∠A + ∠B + ∠BCA = 180

20 + 90 + ∠BCA = 180

∠BCA = 180 - 110

∠BCA = 70

∠BCD = 55

∠ACB + ∠BCD + ∠DCE = 180

70 + 55 + ∠DCE = 180

125 + ∠DCE = 180

Subtract 125 on both sides.

∠DCE = 180 - 125

∠DCE = 55

Problem 5 :

Solution :

∠A + ∠B + ∠ACB = 180

84 + 43 + ∠ACB = 180

127 + ∠ACB = 180

∠ACB = 180 - 127

∠ACB = 53

∠ACB = ∠CDE (vertically opposite angles)

∠CDE + ∠D + ∠E = 180

53 + 97 + ∠E = 180

150 + ∠E = 180

Subtract 150.

∠E = 180 - 150

∠E = 30

Problem 6 :

Solution :

∠DCE + ∠D + ∠E = 180

∠DCE + 35 + 45 = 180

∠DCE + 80 = 180

Subtract 80.

∠DCE = 180 - 80

∠DCE = 100

In triangle ACD.

∠DCE = ∠ACB = 100 (Vertically opposite angles)

∠ACB + ∠A + ∠B = 180

100 + ∠A + 52 = 180

152 + ∠A = 180

Subtracting 152 on both sides.

∠A = 180 - 152

∠A = 38