In the figure, AT is a tangent to a circle with center O. If OT = 4 cm

1.	In the figure, AT is a tangent to a circle with center O. If OT = 4 cm and ∠OTA = 30°, then AT is equals to.(A) 4 cm(B) 2 cm(C) 2\(\sqrt { 3 }\) cm(D) 4\(\sqrt { 3 }\) cm
Answer» Answer is (C) 2\(\sqrt { 3 }\) cm Join OA. since AT is the tangent on the circle an OA is the radius of the circle. and OA is the radius of the circle. So, AT ⊥ OA i.e., ∠OAT = 90° From the right angled ΔOAT ∠OTA = 30° and OT = 4 cm cos 30° = \(\frac { AT }{ OT }\) ⇒ \(\frac { \sqrt { 3 } }{ 2 }\) = \(\frac { AT }{ 4 }\) ⇒ AT = \(\frac { 4\sqrt { 3 } }{ 2 } \) = 2\(\sqrt { 3 }\)

In the figure, AT is a tangent to a circle with center O. If OT = 4 cm and ∠OTA = 30°, then AT is equals to.(A) 4 cm(B) 2 cm(C) 2\(\sqrt { 3 }\) cm(D) 4\(\sqrt { 3 }\) cm

Answer»

Answer is (C) 2\(\sqrt { 3 }\) cm

Join OA. since AT is the tangent on the circle an OA is the radius of the circle. and OA is the radius of the circle.

So, AT ⊥ OA i.e., ∠OAT = 90°

From the right angled ΔOAT

∠OTA = 30° and OT = 4 cm

cos 30° = \(\frac { AT }{ OT }\)

⇒ \(\frac { \sqrt { 3 } }{ 2 }\) = \(\frac { AT }{ 4 }\)

⇒ AT = \(\frac { 4\sqrt { 3 } }{ 2 } \)

= 2\(\sqrt { 3 }\)