JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

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Application +4 | -1

In a

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

and

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

. Then,

a

View answer and explanation

Correct option: B

9 \mathrm{~cm}

Option B (

9 \text{{ cm}}

) is correct because using the area and angle A, we can find bc, and with b-c, we can find a using the cosine rule.

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