SSC CGL 18 Aug 2021 Shift 2 Paper

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Question : 53 of 100

Marks: +1, -0

Point

P

lies outside a circle with centre

O

. Tangents

P A

and

P B

are drawn to meet the circle at

A

and

B

respectively. If

∠ A P B=80^{∘}

, then

∠ O A B

is equal to

$70^{∘}$
$40^{∘}$
$140^{∘}$
$35^{∘}$

Solution:

Given,

In quadrilateral

A P B O

.

∠ O+∠ A+∠ P+∠ B=360^{∘}

⇒ \;\; ∠ O=360^{∘}-90^{∘}-90^{∘}-80^{∘}

⇒ \;\; ∠ O=100^{∘}

\;\text " In "\; △ A O B, ∠ O A B=∠ O B A=x

{ \;\; [ ∵ O A=O B=\;\text " Radius "\;]}

Now,

∠ O A B+∠ O B A+A O B=180^{∘}

⇒ \;\; x+x+100^{∘}=180^{∘}

⇒ \;\; 2 x=80^{∘} ⇒ x=40^{∘}

∴ \;\; ∠ O A B=40^{∘}

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