Question

Given \qquad m \angle AOC = 104^\circm∠AOC=104 ∘ m, angle, A, O, C, equals, 104, degrees \qquad m \angle AOB = 7x + 30^\circm∠AOB=7x+30 ∘ m, angle, A, O, B, equals, 7, x, plus, 30, degrees \qquad m \angle BOC = 9x + 42^\circm∠BOC=9x+42 ∘ m, angle, B, O, C, equals, 9, x, plus, 42, degrees Find m\angle BOCm∠BOCm, angle, B, O, C:

Answer 1

∠BOC = 60°

Explanation:

Given the following angles.

∠AOC=104°

∠AOB = (7x + 30)°

∠BOC= (9x + 42)°

Since all the angles have a common point at O, it can be inferred that;

∠AOC = ∠AOB + ∠BOC

104° = (7x + 30)° + (9x + 42)°

104° = 16x+72

16x = 104-72

16x = 32

x = 32/16

x = 2°

To get ∠BOC:

Since ∠BOC = 9x+42, we will substitute x = 2° into the equation to get the angle ∠BOC

∠BOC = 9(2) + 42

∠BOC = 18+42

∠BOC = 60°