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Op is equal to diameter of circle .Then prove that ABP is an equilateral triangle

Posted by Arsh Raj 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Construction : Join A to B
we have,
OP = diameter
{tex}\Rightarrow {/tex} OQ + QP = diameter
{tex}\Rightarrow {/tex} Radius + QP = diameter
{tex}\Rightarrow {/tex} OQ - PQ = radius
Thus OP is the hypotenuse of right angled {tex}\triangle {/tex}AOP
So, In {tex}\angle A O P , \sin \theta = \frac { A O } { O P } = \frac { 1 } { 2 }{/tex}
{tex}\theta = 30 ^ { \circ }{/tex}
Hence, {tex}\angle A P B = 60 ^ { \circ }{/tex}
Now, In {tex}\triangle A O P{/tex}
AP = AB
So, {tex}\angle P A B = \angle P B A = 60 ^ { \circ }{/tex}
{tex}\therefore \triangle A P B{/tex} is an equilateral triangle.

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