Question

In ΔABC, ∠A = 50° and the external bisectors of ∠B and ∠C meet at O as shown in figure. The measure of ∠BOC is
(a) 40°
(b) 65°
(c) 115°
(d) 140°

Answer 1

Given ABC is a triangle and external bisector of ∠B and ∠C meet at O,

And ∠A=50°

Then Ex ∠PBC=∠A+∠C....1)

Then Ex ∠QCB=∠B+∠A...(2)

∠PBC=2∠OBC ( BO is bisector of angle B)

∠QCB=2∠BCO ( CO is bisector of angle C)

Add (1) anbd (2) we get

∠PBC+∠QCB=∠A+∠C+∠B+∠A

⇒2∠OBC+2∠BCO=∠A+180^0

In triangle ABC

∠A+∠B+∠C=180^0

and ∠A=50^0 (given)

Then ∠2∠OBC+2∠BCO=∠A+180^0

⇒∠OBC+∠BCO= 1/2∠+90^0 = 2/50+90=115^0

ΔBOC∠BOC+∠OBC+∠BCO=180^0

⇒∠BOC=180−115=65^0

Answer 2

Answer:

Ur question is based on a theorem

in which boc=90-1/2(A)

according to question

boc=90-1/2(50)

boc=90-25

boc=65.

Step-by-step explanation: