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