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IN ANY TRIANGLE ABC, THERE’S A POINT P IN THE INTERIOR OF THE TRIANGLE. IT IS GIVEN THAT : AP^2 = BP^2 + CP^2. PROVE THAT ANGLE BPC = 150 DEGREE.

IN ANY TRIANGLE ABC, THERE’S A POINT P IN THE INTERIOR OF THE TRIANGLE. IT IS GIVEN THAT :    AP^2 =  BP^2 + CP^2.      PROVE  THAT  ANGLE BPC = 150 DEGREE. 

Grade:11

1 Answers

Sourabh Singh IIT Patna
askIITians Faculty 2104 Points
8 years ago
Hii
Use trigonometry in order to get the relation.. As triangle is not mentioned and not the postion of point also.Use it anywhere in between the triangular area.

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