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Question
In figure, M is the centre of the circle and seg KL is a tangent segment. If MK = 12, KL = `6sqrt(3)`, then find
(i) Radius of the circle.
(ii) Measures of ∠K and ∠M.
Solution
(i)
Line KL is the tangent to the circle at point L and seg ML is the radius. ......[Given]
∴ ∠MLK = 90° ......(i) [Tangent theorem]
In ∆MLK,
∠MLK = 90°
∴ MK2 = ML2 + KL2 .....[Pythagoras theorem]
∴ 122 = ML2 + `(6sqrt(3))^2`
∴ 144 = ML2 + 108
∴ ML2 = 144 − 108
∴ ML2 = 36
∴ ML = `sqrt(36)`
∴ ML = 6 units ......[Taking square root of both sides]
∴ Radius of the circle is 6 units.
(ii)
We know that,
ML = `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]
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