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.
Sumit Bhattacharya , 10 Years ago
Grade 11
