Q. 83.7( 6 Votes )

# In Fig. 6.8, PB = PD. The value of x is

A. 85°

B. 90°

C. 25°

D. 35°

Answer :

Exterior angle of triangle is equal to sum of 2 opposite onterior angles.

As BC is straight line

∠PBD + 120° = 180°

∠PBD = 180° - 120°

∠PBD = 60°

As ΔPBD is isosceles triangle and PB = PD

∴ ∠PBD = ∠PDB

⇒ ∠PDB = 60°

In ΔPSR

With exterior angle ∠PSQ equal to sum of opposite inetrior angles

∠PDB = ∠DPC + ∠PCD

60° = x + 35°

x = 60° - 35°

x = 25°

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