SSC CHSL 6 Aug 2021 Shift 1 Paper

Section: Quantitative Aptitude

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Question : 72 of 100

Marks: +1, -0

In a

\triangle A B C

, right-angled at

C

if sec

A=\frac{13}{5}

, then find the value of

\frac{1+\sin A}{\cos B}

.

$\;{25}/{12}$
$\;{3}/{2}$
$\;{18}/{5}$
5

Solution:

Given,

\sec A=\;{\;\text " Hypotenuse "\;}/{\;\text " Base "\;}=\;{B A}/{A C}=\;{13}/{5}

In

△ A B C

B C^{2} \;\; =A B^{2}-A C^{2}=(13)^{2}-5^{2}

B C^{2} \;\; =169-25=144

B C \;\; =12

∴ \;\; \sin A \;\; =\;{12}/{13} \;\text " and "\; \cos B=\;{12}/{13}

∴ \;\text " The value of "\;

\;{1+\sin A}/{\cos B}=\;{1+{12}/{13}}/{{12}/{13}}=\;{25}/{13} × \;{13}/{12}=\;{25}/{12}

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