CUET Preparation Today
The tangent at a point A on a circle with center O intersects the diameter PQ of the circle, when extended,at point B. If $\angle BAQ = 105^\circ$, then $\angle APQ$ is equal to:
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$55^\circ$ $60^\circ$ $65^\circ$ $75^\circ$ |
$75^\circ$ |
As we know, \(\angle\)PAQ = 90 [PQ is a diameter] \(\angle\)BAQ = 105 (Given) = \(\angle\)BAP + \(\angle\)PAQ = 105 = \(\angle\)BAP = 105 - 90 = 15 As we know, \(\angle\)BAP = \(\angle\)AQP = 15 If \(\Delta \)APQ \(\angle\)APQ + \(\angle\)PAQ + \(\angle\)AQP = 180 = \(\angle\)APQ + 90 + 15 = 180 Therefore, \(\angle\)APQ = 180 - 90 - 15 = \({75}^\circ\). |
