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m∠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 :

The angle named ∠BOC of the given triangle is calculated as; 60°

How to find the angles in a triangle?

We are given the following angles;

∠AOC=104°

∠AOB = (7x + 30)°

∠BOC= (9x + 42)°

We will notice that all the angles have a common point at O, it can be deduced that;

∠AOC = ∠AOB + ∠BOC

Thus;

104° = (7x + 30)° + (9x + 42)°

104° = 16x + 72

16x = 104 - 72

16x = 32

x = 32/16

x = 2°

Let's now solve for the angle ∠BOC:

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

∠BOC = 9(2) + 42

∠BOC = 18+42

∠BOC = 60°

Read more about Angles in Triangle at; https://brainly.com/question/25215131

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