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,
\[{MK}^2 = {ML}^2 + {LK}^2 \]
\[ \Rightarrow ML = \sqrt{{MK}^2 - {LK}^2}\]
\[ \Rightarrow ML = \sqrt{\left( 12 \right)^2 - \left( 6\sqrt{3} \right)^2}\]
\[ \Rightarrow ML = \sqrt{144 - 108}\]
\[ \Rightarrow ML = \sqrt{36} = 6 \] units
Thus, the radius of the circle is 6 units.

(2)

In right ∆MLK,

\[\tan\angle K = \frac{ML}{KL}\] 

\[ \Rightarrow \tan\angle K = \frac{6}{6\sqrt{3}} = \frac{1}{\sqrt{3}}\] 

\[ \Rightarrow \tan\angle K = \tan30°\] 

\[ \Rightarrow \angle K = 30°\]

Using angle sum property, we have
\[\angle K + \angle L + \angle M = 180^\circ\]
\[ \Rightarrow 30^\circ + 90^\circ + \angle M = 180^\circ\]
\[ \Rightarrow 120^\circ + \angle M = 180^\circ\]
\[ \Rightarrow \angle M = 180^\circ - 120^\circ = 60^\circ\]
Thus, the measures of ∠K and ∠M are 30º and 60º, respectively.

Answer 2

(1)

The line KL is the tangent to the circle at point L and seg ML is the radius.     ...[Given]
∴ ∠MLK = 90º          .... (i) [Tangent theorem]

In right ∆MLK,

∠MLK=90°

\[{MK}^2 = {ML}^2 + {LK}^2 \]          ...[Pythagoras theorem]
\[ \Rightarrow ML = \sqrt{{MK}^2 - {LK}^2}\]
\[ \Rightarrow ML = \sqrt{\left( 12 \right)^2 - \left( 6\sqrt{3} \right)^2}\]
\[ \Rightarrow ML = \sqrt{144 - 108}\]
\[ \Rightarrow ML = \sqrt{36} = 6 \] units.          ...[Taking the square root of both sides]     
Thus, the radius of the circle is 6 units.

(2)

In right ∆MLK,

\[\Rightarrow\mathrm{ML = \frac{1}{2} MK}\]

∴ ∠K = 30°          ...(ii) [Converse of 30° – 60° – 90° theorem]

In ∆MLK,

 ∠L = 90°          ...[From (i)]   

∠K = 30°          ...[From (ii)]

∴ ∠M = 60°          ...[Remaining angle of △MLK]
Thus, the measures of ∠K and ∠M are 30º and 60º, respectively.