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UPSC NDA I 2023 Mathematics Solved Paper

Question Numbers: 35-36

Consider the following for the next two (02) items that follow :

The perimeter of a triangle ABC is 6 times the AM of sine of angles of the triangle.

Further BC = √3 and C

A = 1

Question : 35 of 120

Marks: +1, -0

What is the perimeter of the triangle ?

Solution:

Concept:

Apply the sine rule and the condition relating the perimeter to the arithmetic mean of sines of angles.

Explanation:

Let the sides

a, b, c

be opposite to angles

A, B, C

respectively.

Given:

a = \sqrt{3}

b = 1

Perimeter

P = a + b + c = 6\left(\frac{\sin A + \sin B + \sin C}{3}\right)

\Rightarrow P = 2(\sin A + \sin B + \sin C)

—(1)

By sine rule:

\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = k

Thus

\sin A = ak

\sin B = bk

\sin C = ck

Substitute into (1):

a+b+c = 2k(a+b+c)

\Rightarrow 1 = 2k

\Rightarrow k = \frac{1}{2}

Now

\sin A = a k = \sqrt{3} \times \frac{1}{2} = \frac{\sqrt{3}}{2}

\Rightarrow A = 60^\circ

\sin B = b k = 1 \times \frac{1}{2} = \frac{1}{2}

\Rightarrow B = 30^\circ

Since

A+B+C = 180^\circ

, we get

C = 90^\circ

Now evaluate expression (1):

P = 2(\sin 60^\circ + \sin 30^\circ + \sin 90^\circ)

= 2\left(\frac{\sqrt{3}}{2} + \frac{1}{2} + 1\right) = 2\left(\frac{\sqrt{3}+1+2}{2}\right) = \sqrt{3} + 3

Answer:

3 + \sqrt{3}

(which matches option C:

\sqrt{3} + 3

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