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CDS Model Paper 4 Mathematics

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Question : 60 of 100

Marks: +1, -0

In the adjoining figure

\angle A=60^{\circ}

\angle ABC=80^{\circ}, \angle BQC

$40^{\circ}$
$80^{\circ}$
$20^{\circ}$
$30^{\circ}$

Solution:

A\ B\ C\ D

is a cyclic quadrilateral. Therefore

\angle DCB=180^{\circ}-\angle A=180^{\circ}-60^{\circ}=120^{\circ}

\angle ABC=80^{\circ}

; therefore

\angle BCQ=180^{\circ}-120^{\circ}=60^{\circ}

And

\angle CBQ=180^{\circ}-80^{\circ}=100^{\circ}

(because, sum of angles on a line

=180^{\circ}

)

Then in

\Delta BCQ, \angle Q=180^{\circ}-(100^{\circ}+60^{\circ})=20

(\because

sum of angles of triangle

=180^{\circ})

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