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UPSC NDA Math Model Paper 1

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Question : 26 of 120

Marks: +1, -0

In a

\triangle ABC

a=18, b=24

c=30

then find the value of

\sin\left(\frac{A}{2}\right)

?

$\frac{1}{\sqrt{10}}$
$\frac{1}{\sqrt{5}}$
$\frac{1}{\sqrt{15}}$
None of these

Solution:

Given: For

\triangle ABC

we have

a=18, b=24

and

c=30

Here, we have to find the value of

\sin\left(\frac{A}{2}\right)

As we know that, if

a, b

and

c

are the sides of the

\triangle

ABC then

2S=a+b+c

\Rightarrow 2S=18+24+30=72

\Rightarrow S=36

As we know that,

\sin\frac{A}{2} = \sqrt{\frac{(S-b)(S-C)}{bc}}

\Rightarrow \sin\frac{A}{2} = \sqrt{\frac{(36-24)\times(36-30)}{24\times 30}}

= \frac{1}{\sqrt{10}}

Hence, option

A

is the correct answer.

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