Question

solve these 2 question and answer giver will be marked as brainliest

Answer 1

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here's your answer ⬇⬇

$<b>8.Solution</b>$

PA and PB are the tangents to the circle. ∴ OA ⊥ PA ⇒ ∠OAP = 90° In ΔOPA, sin ∠OPA = OA OP = r 2r [Given OP is the diameter of the circle] ⇒ sin ∠OPA = 1 2 = sin 30 ⁰

⇒ ∠OPA = 30°

Similarly, it can be proved that ∠OPB = 30°.

Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°

In ΔPAB,

PA = PB [lengths of tangents drawn from an external point to a circle are equal]

⇒∠PAB = ∠PBA ............(1) [Equal sides have equal angles opposite to them]

∠PAB + ∠PBA + ∠APB = 180° [Angle sum property]

⇒∠PAB + ∠PAB = 180° – 60° = 120° [Using (1)]

⇒2∠PAB = 120°

⇒∠PAB = 60° .............(2)

From (1) and (2)

∠PAB = ∠PBA = ∠APB = 60°

∴ ΔPAB is an equilateral triangle.

$<b>9.Solution</b>$

Given: O is the centre of the circle. PA and PB are tangents drawn to a circle and ∠APB = 120°.

To prove: OP = 2AP

Proof:

In ΔOAP and ΔOBP,

OP = OP (Common)

∠OAP = ∠OBP (90°) (Radius is perpendicular to the tangent at the point of contact)
OA = OB (Radius of the circle)

∴ ΔOAP is congruent to ΔOBP (RHS criterion)

∠OPA = ∠OPB = 120°/2 = 60° (CPCT)
In ΔOAP,
cos∠OPA = cos 60° = AP/OP
Therefore, 1/2 =AP/OP
Thus, OP = 2AP

Hence, proved.