In {ABC}, with usual notations, if B/2={c+a/2 a}, then a²=

Correct Answer is : b2+c2b²+c²b2+c2

Correct Answer is : b2+c2b²+c²b2+c2

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MHTCET 23 Apr 2025 Shift 1 Paper

Question : 103 of 150

Marks: +1, -0

\triangle \mathrm{ABC}

\cos \frac{B}{2}=\sqrt{\frac{c+a}{2 a}}

a^{2}=

Solution:

Concept:

Use the half-angle formula

\cos \frac{B}{2} = \sqrt{\frac{s(s-b)}{ac}}

, where

s

is the semi-perimeter.

Explanation:

Given:

\cos \frac{B}{2} = \sqrt{\frac{c+a}{2a}}

Equating with the half-angle formula:

\sqrt{\frac{c+a}{2a}} = \sqrt{\frac{s(s-b)}{ac}}

Squaring both sides:

\frac{c+a}{2a} = \frac{s(s-b)}{ac}

Since

s = \frac{a+b+c}{2}

, we get:

\frac{c+a}{2a} = \frac{\left(\frac{a+b+c}{2}\right)\left(\frac{a+c-b}{2}\right)}{ac}

Simplify the numerator using

(a+c+b)(a+c-b) = (a+c)^2 - b^2

\frac{c+a}{2a} = \frac{(a+c)^2 - b^2}{4ac}

Cross-multiplying:

2c(c+a) = (a+c)^2 - b^2

2c^2 + 2ac = a^2 + 2ac + c^2 - b^2

Cancelling

2ac + c^2

from both sides:

a^2 = b^2 + c^2

Answer:

a^2 = b^2 + c^2

This matches option C.

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