In the 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 `.

Explanation

In `Delta MLK`,
`/_ MLK = 90^(@)` ….(Tangent theorem)
`:.` by Pythagoras theorem,
`MK^(2) = ML^(2) + LK^(2)`
`:. 12^(2) = ML^(2) + ( 6 sqrt(3))^(2) `
`:. 144 = ML^(2) + 36 xx 3 `
`:. 144 = ML^(2) + 108`
`:. ML^(2) = 144 - 108`
`:. ML^(2) = 36`
`:. ML= 6` ...(Takin quare roots of both the sides )
`:. ` radius of the circle = ML = 6 .
In `Delta MLK`,
`ML = (1)/(2) MK `
`:. /_ K = 30^(@)` ...(By converse of `30^(@) - 60^(@) - 90^(@)` triangle theorem )
In `Delta MLK`,
`/_ M + /_ K + /_ L = 180^(@)` ...(Sum of all angles of a triangle is `180^(@)0`
`:. /_ M + 30^(@) + 90^(@) = 180^(@0`
`:. /_ M + 120^(@) = 180^(@)`
`:. /_ M = 180^(@) - 120^(@)`
`:. /_ M = 60^(@)`