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PA and PB are tangents drawn to a circle with centre O. If $\angle AOB = 120^\circ$ and $OA = 10 \text{ cm}$, then
(i) Find $\angle OPA$.
(ii) Find the perimeter of $\triangle OAP$.
(iii) Find the length of chord AB.
(i) Find $\angle OPA$.
(ii) Find the perimeter of $\triangle OAP$.
(iii) Find the length of chord AB.
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(i) $\angle OPA = 30^\circ$ (1 mark)
(ii) In $\triangle OAP, \sin 30^\circ = \frac{10}{OP} \Rightarrow OP = 20 \text{ cm}$ (1 mark)
$\tan 30^\circ = \frac{10}{PA} \Rightarrow PA = 10\sqrt{3} \text{ cm}$ (1 mark)
$\therefore \text{Perimeter of } \triangle OPA = (30 + 10\sqrt{3}) \text{ cm}$ (1 mark)
(iii) $PA = PB$ and $\angle APB = 60^\circ \Rightarrow \triangle APB$ is an equilateral triangle ($\frac{1}{2}$ mark)
$\therefore PA = AB = 10\sqrt{3} \text{ cm}$ ($\frac{1}{2}$ mark)
(ii) In $\triangle OAP, \sin 30^\circ = \frac{10}{OP} \Rightarrow OP = 20 \text{ cm}$ (1 mark)
$\tan 30^\circ = \frac{10}{PA} \Rightarrow PA = 10\sqrt{3} \text{ cm}$ (1 mark)
$\therefore \text{Perimeter of } \triangle OPA = (30 + 10\sqrt{3}) \text{ cm}$ (1 mark)
(iii) $PA = PB$ and $\angle APB = 60^\circ \Rightarrow \triangle APB$ is an equilateral triangle ($\frac{1}{2}$ mark)
$\therefore PA = AB = 10\sqrt{3} \text{ cm}$ ($\frac{1}{2}$ mark)