Question

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

Answer 1

In above figure,

Δ ABC is an equilateral triangle and Δ PBC is an isosceles triangle and ∠PBA = 20°.

∵ Δ ABC is an equilateral triangle

∴ ∠ABC = 60°

⇒ ∠PBA + ∠PBC = 60°

⇒ 20° + ∠PBC =  60°

⇒ ∠PBC  = 60° - 20° = 40°

∵ ∠PBC is an isosceles triangle,

∴ ∠PBC = ∠PCB = 40°

Now in ΔBPC,

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

⇒ 40° + 40° + ∠BPC = 180°

⇒ 80° + ∠BPC = 180°

⇒ ∠BPC = 180° - 80° = 100°