Question

In the following figure, triangle ABC is equilateral and triangle PBC is isosceles. If PBA = 20°; find angle BPC.

Answer 1

In the above figure,

∠PBA = ∠ABC + ∠PBA = 60° + 20° = 80°

But ΔPBC is an isosceles triangle

∴ ∠PCB = ∠PBC = 80°

But ∠PBC + ∠PCB + ∠BPC = 180°   ...(Sum of angles of a triangle)

⇒ 80° + 80° + ∠BPC = 180°

⇒ 160° + ∠BPC = 180°

∠BPC = 180° - 160° = 20°