Question

In figure AB is a diameter of a circle with centre O and AT is a tangent. If $\angle AOQ={58}^{\circ }$, find $\angle ATQ$.

Answer 1

$\angle ABQ=\frac{1}{2}\angle AOQ$
$\Rightarrow \frac{1}{2}\times 58=29$
$\angle$ A =90 (AT is a tangent)
$\angle$ BAT+ $\angle$ ABT + $\angle$ ATQ = 180 (angle sum property of triangle)

90 +29 + $\angle$ ATQ = ${180}^{\circ }$
$\angle$ ATQ = 180 - 119
$\angle$ ATQ = ${61}^{\circ }$