In ΔABC, ∠A = 50° and the external bisectors of ∠B and ∠C meet at O as shown in the figure. Find the measure of ∠BOC.

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

And ∠A = 50^o

Then Ex∠PBC = ∠A + ∠C...........................(1)

Ex∠QCB =∠B+∠A...........................(2)

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

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

Add (1) and (2), we get:

⇒ ∠PBC + ∠QCB = ∠A + ∠C + ∠B + ∠A

⇒ 2∠OBC + 2∠BCO = ∠A + 180⁰

In triangle ABC,  ∠A + ∠B + ∠C = 180⁰ and ∠A = 50⁰ .... (given)

Then, 2∠OBC + 2∠BCO = ∠A + 180⁰

⇒ ∠OBC + ∠BCO = $\frac{1}{2}\angle \mathrm{A}$ + 90⁰ 

                               = 25⁰​ + 90⁰ = 115⁰

In ΔBOC,

∠BOC + ∠OBC + ∠BCO = 180⁰

⇒∠BOC = 180⁰ −115⁰ = 65⁰