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From Karnataka Open Educational Resources
Circles Tangents Problems (view source)
Revision as of 05:28, 14 August 2014
, 05:28, 14 August 2014→Algorithm
∴∠QPB=x˚ (∵PQ=BQ)<br>
∴∠QPB=x˚ (∵PQ=BQ)<br>
Now Let ∠PAQ=Y˚<br>
Now Let ∠PAQ=y˚<br>
∠QPA=Y˚ (∵ PQ=AQ)<br>
∠QPA=y˚ (∵ PQ=AQ)<br>
∴In △PAB<br>
∴In △PAB<br>
∠PAB+∠PBA+∠APB=180˚
∠PAB+∠PBA+∠APB=180˚<br>
y˚+x˚+(x˚+y˚)=180˚
2x˚+2y˚=180˚
2(x˚+y˚)=180˚
x˚+y˚=90˚
∴ ∠APB=90˚
=problem 3 [Ex-15.2 B.7]=
=problem 3 [Ex-15.2 B.7]=
