Question

In the given figure, BO and CO are the bisectors of the exterior angles meeting each
other at O. If <A = 70° find <BOC.



please answer fasttttt.

​

Answer 1

Step-by-step explanation:

Angle A = 70°

Angle B + Angle C = 180° - 70° ( ANGLE SUM PROPERTY )

Angle B + Angle C = 110°

Angle B = 110° - Angle C

Angle EBC = 180 - ( 110° - ANGLE C) ( LINEAR PAIR )

Angle EBC = 70° + ANGLE C

SINCE BO IS A BISECTOR

ANGLE OBC = ANGLE EBC / 2

SO ANGLE OBC = 35° + ANGLE C /2

SIMILARLY ANGLE BCO = 35° + ANGLE B/2

ANGLE BOC = 180° - ( 70° + Angle B /2 + Angle C / 2)

[[[[Angle BAC + Angle ABC + Angle ACB = 180°

ANGLE B + ANGLE C = 180° - ( ANGLE A)

DIVIDE BOTH SIDES BY 2

THEREFORE

ANGLE B /2 + ANGLE C /2 = 90° - ANGLE A /2]]]]]

Angle BOC = 110° - ( 90° - Angle A/2)

Angel BOC = 110° - ( 90° - 35° )

Angle BOC = 55°

THEREFORE ANGLE BOC = 55°

Pls mark me brainliest. Thanks

Answer 2

Answer:

I hope it's helpful for you