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Given \qquad \overline{OA}\perp\overline{OC} OA ⊥ OC start overline, O, A, end overline, \perp, start overline, O, C, end overline \qquad m \angle AOB = 6x - 12^\circm∠AOB=6x−12 ∘ m, angle, A, O, B, equals, 6, x, minus, 12, degrees \qquad m \angle BOC = 3x + 30^\circm∠BOC=3x+30 ∘

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Answer:

Step-by-step explanation:

If OA is perpendicular to OC, then <OAC = 90°

<OAC = <AOB + <BOC

Given

<AOB = 6x-12

<BOC = 3x+30

On substituting this values into the formula:

6x-12+3x+30 = 90

9x+18 = 90

Subtract 18 from both sides

9x+18-18 = 90-18

9x = 72

x = 72/9

x = 8°

Since <AOB = 6x-12

<AOB = 6(8)-12

<AOB = 48-12

<AOB = 36°

Also <BOC = 3x+30

<BOC = 3(8)+30

<BOC = 24+30

<BOC = 54°

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