ΔQER is an isosceles triangle and the bisector of ∠Q and ∠R meet at point E. If ∠QER is 110°. Find the value of ∠QPR.

- 20°
- 65°
- 115°
- 40°

Option 4 : 40°

**Given:**

ΔQER is an isosceles triangle.

QE and RE are bisectors of ∠Q and ∠R

∠QER = 110°

**Calculation:**

In ΔQER,

∠QER + ∠ERQ + ∠RQE = 180°

⇒∠ERQ + ∠RQE = 180° - 110°

⇒ 2∠ERQ = 70°

⇒ ∠ERQ = 35°

⇒ ∠ERQ = ∠RQE = 35°

QE and RE are bisectors.

So,

⇒ ∠PQR = 2∠EQR = 2∠PQE

⇒ ∠PRQ = 2∠PRE = 2∠ERQ

Now,

⇒ ∠RQE = ∠ERQ = 35°

⇒ ∠PQR = ∠PRQ = 70°

In ΔQPR,

⇒ ∠QPR + ∠PRQ + ∠RQP = 180°

⇒ ∠QPR = 180° - 140°

∴ ∠QPR = 40°

The correct option is 4 i.e. 40°

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