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:
Answer & explanation
Correct answer: option 4
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 = \({75}^\circ\).