Question

In the ddjoining figure $O$ is the centre of the circle. Forom point $RsegRM$ and $segRN$ are tengent segment touching the circle at $M$ and $N$. If $\left(OR\right)=10cm$ and radius of the circle $=5cm$, then
What is the measure os $\angle MRN$?

Answer 1

We have $OM=5$cm ad $OR=10$cm

In rightangled $△OMR,$

${OR}^2={OM}^2+{MR}^2$

$\Rightarrow MR=\sqrt{{OR}^2-{OM}^2}=\sqrt{{10}^2-5^2}=\sqrt{100-25}=\sqrt{75}=5\sqrt{3}$cm

Tangent segments drawn from an external point to a circle are congruent.

$\therefore MR=NR=5\sqrt{3}$cm

In right angled $△OMR$

$\tan \angle MRO=\frac{OM}{MR}$

$=\frac{5}{5\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \tan \angle MRO=\tan {30}^{\circ }$

Thus, the measure of $\angle MRO$ is ${30}^{\circ }$

In right angled $△ONR$

$\tan \angle NRO=\frac{ON}{NR}$

$=\frac{5}{5\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \tan \angle NRO=\tan {30}^{\circ }$

Thus, the measure of $\angle NRO$ is ${30}^{\circ }$

$\therefore \angle MRN=\angle MRO+\angle NRO={30}^{\circ }+{30}^{\circ }={60}^{\circ }$

Thus, the measure of $\angle MRN$ is ${60}^{\circ }$