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UPSC NDA II 2023 Mathematics Solved Paper

Question Numbers: 79-80

Consider the following for the next items that follow:

The angles A, B and C of a triangle ABC are in the ratio 3 ∶ 5 ∶ 4.

Question : 80 of 120

Marks: +1, -0

What is the ratio of a2 ∶ b2 ∶ c2 ?

Solution:

Concept:

The sine rule relates side lengths to opposite angles in any triangle.

Explanation:

Let the angles

A

B

C

3x

5x

4x

respectively.

Sum of angles =

180^\circ

3x+5x+4x=180 \implies 12x=180 \implies x=15

Thus

A=45^\circ

B=75^\circ

C=60^\circ

By the sine rule:

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

(constant).

a=k\sin45^\circ = k\cdot\frac{1}{\sqrt{2}}

b=k\sin75^\circ

c=k\sin60^\circ = k\cdot\frac{\sqrt{3}}{2}

\sin75^\circ = \sin(45^\circ+30^\circ)=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ = \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\cdot\frac12 = \frac{\sqrt{3}+1}{2\sqrt{2}}

Hence

a^2 = k^2\cdot\frac12

b^2 = k^2\cdot\frac{(\sqrt{3}+1)^2}{8} = k^2\cdot\frac{4+2\sqrt{3}}{8}

c^2 = k^2\cdot\frac{3}{4}

Express all with denominator

8

a^2 = \frac{4k^2}{8}

b^2 = \frac{(4+2\sqrt{3})k^2}{8}

c^2 = \frac{6k^2}{8}

Ratio

a^2:b^2:c^2 = 4: (4+2\sqrt{3}):6

Dividing by

2

2:(2+\sqrt{3}):3

Answer:

2:2+\sqrt{3}:3

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