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In Figure, AT is a tangent to the circle with centre O such that OT = 4 cm and ∠OTA = 30°. Then AT is equal to

(A) 4 cm                      (B) 2 cm                      (C) 2\sqrt{3}cm                  (D) 4\sqrt{3}cm

Answers (1)

Answer(C) 2\sqrt{3}cm

Given ∠OTA = 30°  , OT = 4cm
Join OA, since AT is tangent.                        
Hence it is perpendicular to OA
In \bigtriangleupOAT
\cos \theta = \frac{B}{H}
\cos 30^{\circ} = \frac{AT}{OT}
\frac{\sqrt{3}}{2}= \frac{AT}{4}\; \; \left ( \because \cos 30^{\circ}= \frac{\sqrt{3}}{2} \right )
AT= 4\times \frac{\sqrt{3}}{2}
AT= 2\sqrt{3}cm


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