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​

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°

In triangle ABC

∠A + ∠B + ∠C = 180°

and

∠A = 50° (given)

Then,

∠2∠OBC + 2∠BCO = ∠A+180°

⇒ ∠OBC + ∠BCO = 1/2

=> ∠+90° = 50/2 + 90 = 115°

ΔBOC

∠BOC+∠OBC+∠BCO = 180°

⇒∠BOC=180−115=65°

$\:$

Answer 2

Given :- In ΔABC, ∠A = 50° and the external bisectors of ∠B and ∠C meet at O .

To Find :- ∠BOC = ?

Solution :-

→ Ext.(∠BCD) = ∠A + ∠B { Exterior angle is equal to sum of opposite interior angles . }

So,

→ ∠BCO = (1/2)Ext.(∠BCD) { CO is angular bisector of Ext.(∠BCD) .}

→ ∠BCO = (1/2)[∠A + ∠B] -------- Eqn.(1)

similarly,

→ Ext.(∠CBE) = ∠A + ∠C { same above }

→ ∠CBO = (1/2)Ext.(∠CBE) { BO is angular bisector . }

→ ∠CBO = (1/2)[∠A + ∠C] --------- Eqn.(2)

Now, in ∆BOC we have,

→ ∠BCO + ∠CBO + ∠BOC = 180° { By angle sum property. }

putting values from Eqn.(1) and Eqn.(2),

→ (1/2)(∠A + ∠B) + (1/2)(∠A + ∠C) + ∠BOC = 180°

→ (1/2)(∠A + ∠A + ∠B + ∠C) + ∠BOC = 180°

putting ∠A + ∠B + ∠C = 180° ,

→ (1/2)(∠A + 180°) + ∠BOC = 180°

→ (1/2)∠A + 90° + ∠BOC = 180°

→ ∠BOC = 180° - 90° - (1/2)∠A

→ ∠BOC = [90° - (1/2)∠A]

therefore,

→ ∠BOC = 90° - (1/2) × 50°

→ ∠BOC = 90° - 25°

→ ∠BOC = 65° (Ans.)

Hence, ∠BOC is equal to 65° .

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