Question

In the figure, O is the center of the circle, PQ is tangent to the circle at A. If $\angle PAB={58}^{\circ }$, find $\angle ABQ$ and $\angle AQB$. [2 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark

Join OA

As the tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\angle OAP={90}^{\circ }$

$\angle OAB=\angle OAP-\angle PAB$

$\angle OAB={90}^{\circ }-{58}^{\circ }$

$\angle OAB={32}^{\circ }$

OA = OB (Radius)

$\angle OAB=\angle OBA$

Therefore $\angle OBA={32}^{\circ }$

$\angle AOQ=\angle OAB+\angle OBA$ (By exterior angle property)

$={32}^{\circ }$ + ${32}^{\circ }$

$={64}^{\circ }$

In $△OAQ$

By angle sum property

$\angle OQA={180}^{\circ }-\angle AOQ-\angle OAQ$

$\angle OQA={180}^{\circ }-{64}^{\circ }-{90}^{\circ }$

$\angle OQA={26}^{\circ }$

Therefore $\angle ABQ={32}^{\circ }$ and $\angle AQB={26}^{\circ }$