In the figure if ∠APB = 60° and OP = 10 cm, then PA = ………cm(A

1.	In the figure if ∠APB = 60° and OP = 10 cm, then PA = ………cm(A) 5 (B) 5√2 (C) 5√3 (D) 20
Answer» Correct option is: (C) 5√3 \(\because\) \(\angle\) APB = 60° = \(\angle\) APO = \(\frac {\angle APB}2 = \frac {60^\circ}{2}\) = 30° Also OA \(\perp\) AP (Angle between radius and tangent at point of contact) \(\therefore\) \(\angle\) OAP = 90° Now, in right \(\triangle\) OAP \(\frac {PA}{OP}\) = cos (\(\angle\)APO) = \(\frac {PA}{10}\) = cos 30° (\(\because\) \(\angle\) OPA = 30° & OP = 10 cm). = PA = 10 cos 30° = \(\frac {\sqrt3}2 \times 10 = 5\sqrt3\). Correct option is: (C) 5√3

In the figure if ∠APB = 60° and OP = 10 cm, then PA = ………cm(A) 5 (B) 5√2 (C) 5√3 (D) 20

Answer»

Correct option is: (C) 5√3

\(\because\) \(\angle\) APB = 60°

= \(\angle\) APO = \(\frac {\angle APB}2 = \frac {60^\circ}{2}\) = 30°

Also OA \(\perp\) AP (Angle between radius and tangent at point of contact)

\(\therefore\) \(\angle\) OAP = 90°

Now, in right \(\triangle\) OAP

\(\frac {PA}{OP}\) = cos (\(\angle\)APO)

= \(\frac {PA}{10}\) = cos 30° (\(\because\) \(\angle\) OPA = 30° & OP = 10 cm).

= PA = 10 cos 30° = \(\frac {\sqrt3}2 \times 10 = 5\sqrt3\).

Correct option is: (C) 5√3

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In the figure if ∠APB = 60° and OP = 10 cm, then PA = ………cm(A) 5 (B) 5√2 (C) 5√3 (D) 20

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