Question

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

∘

m, angle, B, O, C, equals, 3, x, plus, 30, degrees

Find m\angle AOBm∠AOBm, angle, A, O, B:

Answer 1

its 36 degrees i just did it

Explanation:

Answer 2

36°

Explanation:

We know that angles BOC and AOB are complementary angles, that is, the sum 90°. Additionally, we know that

`\angle BOC=3x+30` and
`\angle AOB = 6x-12`, which can be expressed as

`3x+30+6x-12=90`

Let's solve for
`x`

`9x=72\\x=(72)/(9)\\ x=8`

Using this vale, we find the measure of angle AOB

`\angle AOB = 6x-12=6(8)-12 =48-12=36\°`

Therefore, the answer is 36°.