Question

In figure 3.83, M is the centre of the circle and seg KL is a tangent segment. If MK=12,KL= 6√3 then find -
(1) Radius of the circle.
(2) Measures of ∠K and ∠M.

Answer 1

Answer:

1) . Radius of the circle = 6

2). $\angle M = 60^0, K = 30^0$

Step-by-step explanation:

We have given:

MK =12, KL = 6$\sqrt{3}$,

First we find radius of the circle.

In right angle triangle MLK,

We will apply pythagorean theorem here,

$h^2 = b^2 + p^2\\MK^2 = LK^2 + ML^2 \\12^2 = (6\sqrt{3} )^2 + ML^2\\144 = 108 + ML^2\\144 - 108 = ML^2\\ML^2 = 36\\ML = \sqrt{36}\\ ML = 6$

1) . Radius of the circle = 6

Now we will find the angles

$tan K = \dfrac {ML}{LK}\\tan K = \dfrac 6{6\sqrt{3} }\\tanK = \dfrac {1}{\sqrt{3} }\\tan K = tan 30^0\\K = 30^0$

By using angle sum property,

We know that sum of all angles are equal.

$\angle K + \angle L + \angle M = 180^o\\30^o + 90^0 + \angle M = 180^0\\\angle M = 180^0 - (30^o + 90^0)\\\angle M = 60^0$

2). $\angle M = 60^0, K = 30^0$

Answer 2

1) Radius of the circle

Step-by-step explanation:

The tangent at any point of is a circle perpendicular to the radius though the point of contact

Angla MLK =90°

In right angle ∆ MLK

MK²=ML²+LK²

ML²=√Mk²-LK²

ML²=√12²-(6√3

ML²=√144-108

ML²=36

ML ²=6