SSC CHSL 10 Aug 2021 Shift 1 Paper

Section: Quantitative Aptitude

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Question : 61 of 100

Marks: +1, -0

Let

O

be the centre of a circle and

A C

be the diameter.

B D

is a chord intersecting

A C

at

E . A D

and

A B

are joined. If

∠ B O C=40^{∘}

and

∠ A O D=120^{∘}

, then

∠ B E C

is equal to

$55^{∘}$
$80^{∘}$
$90^{∘}$
$70^{∘}$

Solution:

Given,

Since,

∠ A O D=120^{∘}

∴ \;\; ∠ A B D=\;{120^{∘}}/{2}=60^{∘}

Since,

∠ B O C=40^{∘}

∴ \;\; ∠ A O B=180^{∘}-40^{∘}=140^{∘}

Since,

\;\; A O=B O \;\; [ ∵

radii

]

∴

Opposite angles are equal (let

x

).

∴ \;\; ∠ A B O=x=\;{180^{∘}-140^{∘}}/{2}=20^{∘}

∴ \;\; ∠ O B E=60^{∘}-20^{∘}=40^{∘}

[ ∵ ∠ A B D-∠ A B O]

∴ \ln △ B O E, ∠ B E C=∠ B O E+∠ O B E

=40^{∘}+40^{∘}=80^{∘}

[ ∵

external angle

]

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