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

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

Given:
PA and PB are tangents to a circle with center O.
∠AOB = 120°, OA = OB = 10 cm
Since PA and PB are tangents, they are perpendicular to the radii at the points of tangency.

(a) Find ∠OPA:

Since the tangents from an external point are equal, PA = PB.
Triangle OAP is an isosceles triangle with OA = OP.
∠OAP = ∠OPB.

In triangle AOB, the exterior angle theorem states:
∠OPA = (180° - ∠AOB) / 2 = (180° - 120°) / 2 = 60° / 2 = 30°.

Thus, ∠OPA = 30°.

(b) Find the perimeter of triangle OAP:

We know that OA = OP = 10 cm.
Since PA is a tangent, we use the sine rule:
PA = OA × sin ∠OPA = 10 × sin 30° = 10 × 0.5 = 5 cm.

Perimeter of triangle OAP = OA + OP + PA = 10 + 10 + 5 = 25 cm.

(c) Find the length of chord AB:

In triangle AOB, drop a perpendicular from O to AB at M.
OM bisects AB because OA = OB.
∠AOM = ∠AOB / 2 = 120° / 2 = 60°.

Using trigonometry:
AM = OA × sin ∠AOM = 10 × sin 60° = 10 × √3 / 2 = 5√3 cm.

AB = 2 × AM = 2 × 5√3 = 10√3 cm.

Final Answers:

1. ∠OPA = 30°
2. Perimeter of triangle OAP = 25 cm
3. AB = 10√3 cm

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