SSC CGL 24 Aug 2021 Shift 3 Paper

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Question : 73 of 100

Marks: +1, -0

In circle with centre

O

, a diameter

A B

is produced to a point

P

lying outside the circle and

P T

is a tangent to the circle at a point

C

on it. If

∠ B P T=28^{∘}

, then what is the measure of

∠ B C P

?

$62^{∘}$
$45^{∘}$
$28^{∘}$
$31^{∘}$

Solution:

Given,

∠ B P T=28^{∘}, P T=

tangent,

A B=

Diameter
In

△ O C P

,

∠ O C P+∠ C P O+∠ C O P=180^{∘}

90^{∘}+28^{∘}+∠ C O P=180^{∘}

[ ∵ ∠ O C P=90^{∘}]

⇒ \;\; ∠ C O P=180^{∘}-90^{∘}-28^{∘}

∠ C O P=62^{∘}

where

O C

is perpendicular to tangent

P T

.
In

△ O C B, O C=O B=

radius
so,

\;\; ∠ O C B=∠ O B C

⇒ ∠ O C B+∠ C B O+∠ B O C=180^{∘}

⇒ 2 ∠ O C B=180^{∘}-∠ B O C

=180^{∘}-62^{∘}

⇒ \;\; 2 ∠ O C B=118^{∘}

⇒ \;\; ∠ O C B=59^{∘}

In

△ O C P, \;\; ∠ O C B+∠ B C P=∠ O C P

⇒ \;\; 59^{∘}+x^{∘}=90^{∘}

⇒ \;\; x^{∘}=90^{∘}-59^{∘}

x^{∘}=31^{∘}=∠ B C P

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