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MHTCET 26 Apr 2025 Shift 1 Paper

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Question : 121 of 150

Marks: +1, -0

In a triangle ABC , with usual notations,

(a+b+c)(a+b-c)=3 \mathrm{ab}

\angle C=

$\frac{\pi}{2}$
$\frac{\pi}{4}$
$\frac{\pi}{3}$
$\frac{\pi}{6}$

Solution:

Concept:

Use the cosine rule:

\cos C = \frac{a^2+b^2-c^2}{2ab}

.

Explanation:

Given

(a+b+c)(a+b-c)=3ab

.

Rewrite as

(a+b)^2-c^2=3ab

.

Expanding gives

a^2+b^2+2ab-c^2=3ab

.

Rearranging:

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

.

Divide both sides by

2ab

:

\frac{a^2+b^2-c^2}{2ab}=\frac12

.

By the cosine rule,

\cos C = \frac12

.

Therefore

C=\cos^{-1}\left(\frac12\right)=\frac{\pi}{3}

.

Answer:

\angle C=\frac{\pi}{3}

, which is option C.

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