JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

1 of 1

Application +4 | -1

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

, where

\Delta

is the area of

\Delta \;ABC

, then

\tan A

is equal to

View answer and explanation

Correct option: C

\frac{8}{{15}}

Option C (

\frac{8}{{15}}

) is correct because by using the given condition

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

and trigonometric identities, we can find that

\tan A = \frac{8}{{15}}

More Questions from PROPERTIES OF TRIANGLES

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{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}

A

2 s

\triangle A B C, r+r_1=r_2+r_3

\angle B>\frac{\pi}{3}