SSC CGL 24 Aug 2021 Shift 2 Paper

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Question : 57 of 100

Marks: +1, -0

In a circle with centre

O, A B

is a chord of length

10 {cm}

. Tangents at points

A

and

B

intersect outside the circle at

P

. If

O P=2 O A

, then find the length (in

{cm})

of

A P

.

$12.5$
10
12
15

Solution:

Given,

A B=10 {cm}

,
Let

O B=O A=r, O P=2 O A=2 r

△ O A P

is right angle at

A

.

O A^{2}+A P^{2}=O P^{2}

{[ ∵ (\;\text " Base "\;)^{2}+(\;\text " Perpendicular "\;)^{2}}

=(\;\text " Hypotenuse "\;)^{2}]

(A P)^{2}=(O P)^{2}-(O A)^{2}

\;\; \;\; =(2 r)^{2}-(r)^{2}=4 r^{2}-r^{2}

⇒(A P)^{2}=3 r^{2}

A P=r √{3}

. . . (i)

\;\text " In "\; △ O A P, \tan θ=\;{O A}/{A P}=\;{r}/{r √{3}}

⇒ \tan θ=\;{1}/{√{3}}=\tan 30^{∘}

θ=30^{∘}

∠ O A P+∠ A P O+∠ P O A=180^{∘}

90^{∘}+30^{∘}+∠ P O A=180^{∘}

∠ P O A \;\text " or "\; α=180^{∘}-120^{∘}=60^{∘}

\sin α=\;{A T}/{A O}=\;{5}/{r}

\sin 60^{∘}=\;{5}/{r} ⇒ \;{√{3}}/{2}=\;{5}/{r}

r=\;{10}/{√{3}}

From Eq. (i),

A P=r √{3}=\;{10}/{√{3}} × √{3}=10 {cm}

A P=10 {cm}

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