Question

In the given figure, M is the centre of the circle and seg KL is a tangent segment.
If MK = 12, KL = $6\sqrt{3}$ then find –
(1) Radius of the circle.
(2) Measures of ∠K and ∠M.

Answer 1

(1)
The tangent at any point of a circle is perpendicular to the radius through the point of contact.

∴ ∠MLK = 90º



In right ∆MLK,

$$\begin{gathered} {\mathrm{MK}}^2={\mathrm{ML}}^2+{\mathrm{LK}}^2 \\ \Rightarrow \mathrm{ML}=\sqrt{{\mathrm{MK}}^2-{\mathrm{LK}}^2} \\ \Rightarrow \mathrm{ML}=\sqrt{{\left(12\right)}^2-{\left(6\sqrt{3}\right)}^2} \\ \Rightarrow \mathrm{ML}=\sqrt{144-108} \\ \Rightarrow \mathrm{ML}=\sqrt{36}=6\mathrm{cm} \end{gathered}$$
Thus, the radius of the circle is 6 cm.

(2)
In right ∆MLK,

$$\begin{gathered} \tan \angle \mathrm{K}=\frac{\mathrm{ML}}{\mathrm{KL}} \\ \Rightarrow \tan \angle \mathrm{K}=\frac{6}{6\sqrt{3}}=\frac{1}{\sqrt{3}} \\ \Rightarrow \tan \angle \mathrm{K}=\tan 30° \\ \Rightarrow \angle \mathrm{K}=30° \end{gathered}$$
Using angle sum property, we have

$$\begin{gathered} \angle \mathrm{K}+\angle \mathrm{L}+\angle \mathrm{M}=180° \\ \Rightarrow 30°+90°+\angle \mathrm{M}=180° \\ \Rightarrow 120°+\angle \mathrm{M}=180° \\ \Rightarrow \angle \mathrm{M}=180°-120°=60° \end{gathered}$$
Thus, the measures of ∠K and ∠M are 30º and 60º, respectively.