SSC CGL 10 Jun 2019 Shift 1 Paper

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Question : 62 of 100

Marks: +1, -0

In a

\ΔA B C

right angles at

B, A B=7 c m

and

(A C-B C)

=1 \ {\cm} .

The value of

(\sec C+\cot A)

is:

$19/24$
$4/3$
$3/4$
1

Solution:

\ {\Now}, \sec \ {C}+\cot \ {A}=\ {\ {H}}/{\ {B}}+\ {\ {B}}/{ {P}}

=\ {25}/{24}+\ {7}/{24}=\ {4}/{3}

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