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