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# How to solve?

Let $$\triangle{ABC}$$ be equilateral. $$P$$ is a point in the triangle such that $$AP, BP, CP = 2, 2\sqrt{3},4$$ respectively. Find $$\angle{BPC}$$

Note by Victor Loh
4 years ago

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## Comments

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Let $$X,Y,Z$$ be the reflections of $$P$$ w.r.t. $$AB,BC,CA$$. Then $$XYZ$$ is a triangle with side lengths $$a, \sqrt{3}a, 2a$$ where $$a=2\sqrt{3}$$. It's obvious that $$XYZ$$ is a right triangle with $$\angle{XYZ}=30^{\circ}$$. Hence $$\angle{BPC}=\angle{BYC} = \angle{BYX} + \angle{XYZ} + \angle{ZYC} = 30^{\circ}+30^{\circ}+30^{\circ}=90^{\circ}$$.

- 4 years ago

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One way can be finding the area of the $$ΔABC$$ considering each side of the equilateral triangle be $$x$$ using Heron's formula and then finding the area of the three small triangles inside $$ΔABC$$ in a similar way and then comparing them we get value of $$x$$.Now using the formula $$1/2 \times BP \times PC \times \sin \angle BPC=ar(ΔBPC)$$ and solving for $$\sin \angle BPC$$ we get our answer.

- 4 years ago

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