JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

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In a

\triangle A B C, A=\frac{2 \pi}{3}, b-c=3 \sqrt{3}

and ar

(\triangle A B C)=\frac{9 \sqrt{3}}{2} \mathrm{~cm}^2

. Then a is

View answer and explanation

Correct option: B

9 \mathrm{~cm}

Option B (

9 \mathrm{~cm}

.) is correct because by using the area and the given angle, we can find bc, and then using the cosine rule, we can find a.

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