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Question

From an external point $$P$$, tangents $$PA$$ and $$PB$$ are drawn to a circle with centre $$O$$. If $$\angle PAB = 50^\circ ,$$ then find $$\angle AOB$$.


Solution

Given, $$PA=PB$$, and $$\angle PAB=50^o$$

To prove: Measure of $$\angle AOB$$

Proof: $$PA=PB$$ ($$\because$$ given)

$$\angle PBA =\angle PAB$$ ($$\because$$ angle opposite sides are equal)

$$\therefore \angle PBA=50^o$$.

In $$\Delta PAB$$

$$\angle PBA+\angle PAB+\angle APB=180^o$$ ($$\because$$ A.S.P)

$$\angle APB=180^o-100^o$$

$$\angle APB=80^o$$

$$\angle AOB+\angle APB=180^o$$ ($$\therefore$$ supplementary angles)

$$\angle AOB=180^o-80^o$$

$$\angle AOB=100^o$$

$$\therefore$$ Hence proved.

1220205_1448546_ans_5a977ea2504641efbda2d745c69b1652.jpg

Mathematics

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