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From a point P two tangents are drawn to a circle with center o. if OP = diameter of the circle, show `triangleAPB` is equilateral. |
Answer» In `triangle OAP` OP=2r `Sintheta=(OA)/(OP)=r/(2r)=1/2` `sin theta=30^0` `angleOPA=30^o` similarly, `angleOPB=30^o` `angleAPB=angleOPA+angleOPB` `=30^o+30^o` `=60^o` In`trianglePAB` `anglePAB=anglePBA` `anglePAB+anglePBA+angleAPB=180^o` `2anglePAB=120^o` `angle PAB=60^o` so, `anglePBA=60^o` and`angleAPB=60^o` so, `triangleAPB` is an equilateral triangle |
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