In a $\Delta \,ABC$ if $c = \left( {a + b} \right)\sin \,\,\theta $ and $\cos \,\,\theta = \frac{{k\,\sqrt {ab} }}{{a + b}}$, then $k = $

JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

1 of 1

Analyzing +4 | -1

In a

\Delta \,ABC

c = \left( {a + b} \right)\sin \,\,\theta

and

\cos \,\,\theta = \frac{{k\,\sqrt {ab} }}{{a + b}}

, then

k =

View answer and explanation

Correct option: A

2\,\,\cos \,\,\frac{C}{2}

Option A (

2 \cos \frac{C}{2}

) is correct because using sine rule and half angle formula, we can derive that

k = 2 \cos \frac{C}{2}

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