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In the adjoining figure, M is the center of the circle and seg KL is a tangent segment. If MK = 12, KL = 6√3 ,then find 

i. Radius of the circle. 

ii. Measures of ∠K and ∠M.

1 Answer

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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 + (6√3)2 

∴ 144 = ML2 + 108 

∴ ML2 = 144 – 108 

∴ ML2 = 36 

∴ ML = √36 = 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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