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UPSC NDA I 2023 Mathematics Solved Paper

Question Numbers: 45-46

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

In a triangle POR, P is the largest angle and cosP =

\frac{1}{3}

. Further the in-circle of the triangle touches the sides PQ, QR and RP at N, L and M respectively such that the lengths PN, QL and RM are n, n + 2, n + 4 respectively where n is an integer.

Question : 46 of 120

Marks: +1, -0

What is the length of the smallest side ?

Validate

Solution:

Concept:

Use the cosine rule to relate the sides of a triangle to the cosine of an angle.

Explanation:

Given

\cos P = \frac{1}{3}

and sides

PQ

PR

QR

expressed in terms of

n

Apply the cosine rule:

\cos P = \frac{PQ^2 + PR^2 - QR^2}{2 \cdot PQ \cdot PR}

Substitute:

PQ = 2n+2

PR = 2n+4

QR = 2n+6

\frac{1}{3} = \frac{(2n+2)^2 + (2n+4)^2 - (2n+6)^2}{2(2n+2)(2n+4)}

Simplify numerator:

(4n^2+8n+4) + (4n^2+16n+16) - (4n^2+24n+36) = 4n^2 + 0n - 16 = 4n^2 - 16

Denominator:

2(4n^2+12n+8) = 8n^2+24n+16

Thus

\frac{1}{3} = \frac{4n^2 - 16}{8n^2+24n+16}

Cross-multiply:

8n^2+24n+16 = 12n^2 - 48

Rearrange:

4n^2 - 24n - 64 = 0

→ divide by 4:

n^2 - 6n - 16 = 0

Factor:

(n-8)(n+2)=0

→

n=8

(positive length).

Sides:

PQ = 2(8)+2 = 18

PR = 2(8)+4 = 20

QR = 2(8)+6 = 22

The smallest side is

18

Answer:

18

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