Question

In the adjoining figure, $M$ is the centre of the circle and seg $KL$ is a tangent segment. If $MK=12$, $KL=6\sqrt{3}$, then find
i. Radius of the circle.
ii. Measures of $\angle K$ and $\angle M$.

Answer 1

i. Line $KL$ is the tangent to the circle at point $L$ and seg $ML$ is the radius. [Given] $\therefore \angle MLK={90}^{\circ }\dots \dots \dots \dots$ (i) [Tangent theorem]
In $△MLK,\angle MLK={90}^{\circ }$
$\therefore {MK}^2={ML}^2+{KL}^2$ [Pythagoras theorem]
$\therefore {12}^2={ML}^2+(6\sqrt{3}{)}^2$
$\therefore 144={ML}^2+108$
$\therefore {ML}^2=144-108$
$\therefore {ML}^2=36$
$\therefore ML=\sqrt{36}=6$ units. [Taking square root of both sides]
$\therefore$ Radius of the circle is 6 units.

ii. We know that,
$ML=\frac{1}{2}MK$ $\therefore \angle K={30}^{\circ }$
(ii) [Converse of ${30}^{\circ }-{60}^{\circ }-{90}^{\circ }$ theorem]
In $△MLK$,
$\angle L={90}^{\circ }$ [From (i)]
$\angle K={30}^{\circ }$ [From (ii)]
$\therefore \angle M={60}^{\circ }$ [Remaining angle of $△MLK$ ]