In a $\Delta \,ABC,\,\,A = \frac{{2\;\pi }}{3},\,\,b - c = 3\,\sqrt 3 \;{\rm{cm}}$ and $\Delta = \frac{{9\,\sqrt 3 }}{2}\,\,{\rm{c}}{{\rm{m}}^2}$. Then, $a = $

JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

1 of 1

Application +4 | -1

In a

\Delta \,ABC,\,\,A = \frac{{2\;\pi }}{3},\,\,b - c = 3\,\sqrt 3 \;\mathrm{cm}

and

\Delta = \frac{{9\,\sqrt 3 }}{2}\,\,\mathrm{c}{\mathrm{m}^2}

. Then,

a =

View answer and explanation

Correct option: B

9\;\,\mathrm{cm}

Option B,

9\;\,\mathrm{cm}

, is correct because it is the length of side 'a' calculated using the cosine rule and the given area and angle.

Practice All Chapter MCQs →

{p_1},{p_2},{p_3}

\Delta \;ABC

A,B,C

\Delta

p_1^{ - 2} + p_2^{ - 2} + p_3^{ - 2}

5k,\;6k,\;5k

k

\Delta \,ABC

a = \left( {b - c} \right)\sec \theta

\frac{{2\,\sqrt {bc} }}{{b - c}}\,\,\sin \,\,\frac{A}{2} =

\Delta \;ABC,\frac{{b + c}}{a}

\Delta = {a^2} - {\left( {b - c} \right)^2}

\Delta

\Delta \;ABC

\tan A

A

2 s