Question

In ∆ABC, angle A = 50° and the external bisector angleB and angle C meet at O as shown in the figure. The measure of angle BOC is?​

Answer 1

Answer:

Here is your answer!

Step-by-step explanation:

boc=90-1/2(A)

according to question

boc=90-1/2(50)

boc=90-25

boc=65.

Answer 2

Solution :-

given that,

→ ∠A = 50°

Also, the external bisector ∠B and ∠C meet at O .

So,

→ ∠BOC = 90° - (1/2)∠A

→ ∠BOC = 90° - (1/2)50°

→ ∠BOC = 90° - 25°

→ ∠BOC = 65° (b) (Ans.)

P r o o f :- ∠BOC = 90° - (1/2)∠A .

Refer to this l i n k :- In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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