Question 1791
CBSE Class 10 Mathematics · Circles · Tangents from an External Point
Two parallel tangents touch a circle on opposite sides. A third tangent meets them at \(A\) and \(B\) and touches the circle at \(C\). Prove that \(\angle AOB=90^\circ\), where \(O\) is the centre.
Solution to Question 1791
Complete workingFrom the external point \(A\), the line \(AO\) bisects the angle between the two tangents through \(A\). Similarly, \(BO\) bisects the angle between the tangents through \(B\).
Because the first two tangents are parallel, the interior angles made with transversal \(AB\) are supplementary. Let them be \(2\alpha\) and \(2\beta\). Then
\[\begin{aligned}2\alpha+2\beta &= 180^\circ\Rightarrow\alpha+\beta\\&= 90^\circ.\end{aligned}\]
But \(\angle OAB=\alpha\) and \(\angle ABO=\beta\). Hence (\angle AOB=180^\circ-\(\alpha+\beta\)=90^\circ).
Final answer
\(\angle AOB=90^\circ\).