In a ${\rm{\Delta }}\,ABC,\,\,a = 2b$ and $A = 3B$, the $A = $

JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

1 of 1

Application +4 | -1

In a

\mathrm{\Delta }\,ABC,\,\,a = 2b

and

A = 3B

, the

A =

View answer and explanation

Correct option: A

90^\circ

Option A (

90^\circ

) is correct because by applying the sine rule and trigonometric identities, we find that A = 90 degrees.

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