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In the diagram, square ABCD has sides of length 4, and triangle ABE is equilateral. Line segments BE and AC intersect at P. Point Q is on BC so that PQ is perpendicular to BC and PQ=x.


Determine the measure of angle BPC

 Feb 24, 2021
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There is no diagram, so I am assuming $E\in ABCD$. 

We have $\angle PAE=60-\angle BAC=60-45=15$

And $\angle APE=180-60-15=180-75=105$.

By Vertical angles, $\angle APE=\angle BPC=\boxed{105^{\circ}}$.

 Feb 24, 2021

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