Question

PA and PB are tangents drawn to a circle with centre O. If \(\angle AOB = 120^\circ\) and OA = 10 cm, then

(i) Find \(\angle OPA\).
(ii) Find the perimeter of \(\triangle OAP\).
(iii) Find the length of chord AB.

Answer 1

Problem: Tangents and Triangle Geometry

Given: PA and PB are tangents from an external point P to a circle centered at O. Radius \( OA = OB = 10 \, \text{cm} \), and \( \angle AOB = 120^\circ \).

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(i) Find \( \angle OPA \)

• \( \angle OAP = \angle OBP = 90^\circ \) (radius ⟂ tangent)
• In quadrilateral PAOB: \[ \angle APB + \angle OAP + \angle AOB + \angle OBP = 360^\circ \] \[ \angle APB + 90^\circ + 120^\circ + 90^\circ = 360^\circ \Rightarrow \angle APB = 60^\circ \]
• Since \( \triangle OAP \cong \triangle OBP \), \( \angle OPA = \frac{1}{2} \angle APB = \frac{1}{2} \times 60^\circ = 30^\circ \)
• Answer: \( \boxed{\angle OPA = 30^\circ} \)

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(ii) Find the Perimeter of \( \triangle OAP \)

Given: \( OA = 10 \, \text{cm} \), \( \angle OPA = 30^\circ \), right-angled triangle.

• Using trigonometry: \[ \tan 30^\circ = \frac{OA}{PA} \Rightarrow \frac{1}{\sqrt{3}} = \frac{10}{PA} \Rightarrow PA = 10\sqrt{3} \] \[ \sin 30^\circ = \frac{OA}{OP} = \frac{1}{2} \Rightarrow OP = 20 \]
• Perimeter: \[ OA + PA + OP = 10 + 10\sqrt{3} + 20 = \boxed{30 + 10\sqrt{3} \, \text{cm}} \]

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(iii) Find the Length of Chord AB

In triangle \( \triangle AOB \):

• Using the cosine rule: \[ AB^2 = OA^2 + OB^2 - 2(OA)(OB)\cos(\angle AOB) \] \[ AB^2 = 100 + 100 - 200 \cdot \cos(120^\circ) \] \[ \cos(120^\circ) = -\cos(60^\circ) = -\frac{1}{2} \Rightarrow AB^2 = 200 + 100 = 300 \Rightarrow AB = \sqrt{300} = \boxed{10\sqrt{3}} \]
• Alternate Method: Drop perpendicular OM from O to chord AB.
  • \( \angle AOM = 60^\circ \)
  • \( \sin 60^\circ = \frac{AM}{OA} = \frac{\sqrt{3}}{2} = \frac{AM}{10} \Rightarrow AM = 5\sqrt{3} \)
  • \( AB = 2 \cdot AM = 2 \cdot 5\sqrt{3} = \boxed{10\sqrt{3}} \)