Question

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

(Please give a Detailed Answer)

Answer 1

∠BOC= 65° in the above quadrilateral ABOC.

Step-by-Step Explanation:

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

And ∠A=50°

To Find: ∠BOC

Then Ex ∠PBC=∠A+∠C          -(1)

Then 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+180°

⇒2(∠OBC+∠BCO)=∠A+180°     [∵ ∠A+∠B+∠C=180°]

⇒(∠OBC+∠BCO)=(∠A+180°)/2

⇒∠OBC+∠BCO=   1/2∠A+90°  

⇒∠OBC+∠BCO= 25° + 90°= 115°  -(3)

In ΔBOC,

∠BOC+∠OBC+∠BCO=180°  [∵ In a triangle the sum of all angles is equal to 180°]

Using (3),

⇒∠BOC+(∠OBC+∠BCO)=180°

⇒∠BOC+ 115°=180°

⇒∠BOC =180°- 115°

⇒∠BOC= 65°