In the given figure, ABC is a triangle, prove that angleBPC = 90° +1/2× angleBAC
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angle BAC = 180°-(2theta +2a)
= 180° -2(theta +a)
=2[90°-(theta +a)]
angle BPC = 180°-(theta + a)
= 90°+90°-(theta +a)
= 90° + ½*2[90°-(theta +a)]
= 90° +½angle BAC
PROVED.
Answered by
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Answer:
Step-by-step explanation:
angle BAC = 180°-(2theta +2a)
= 180° -2(theta +a)
=2[90°-(theta +a)]
angle BPC = 180°-(theta + a)
= 90°+90°-(theta +a)
= 90° + ½*2[90°-(theta +a)]
= 90° +½angle BAC
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