JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

1 of 1

Application +4 | -1

O

is the origin and

P\left( {2,3} \right)

and

Q\left( {4,5} \right)

are two points, then

OP.OQ\cos \angle POQ =

View answer and explanation

Correct option: D

Option D (

{\left( {\frac{{{c^2} - {a^2}}}{{2ca}}} \right)^2}

) is correct because by using the trigonometric identity for

\sin 3B

and the given condition

2b^2 = a^2 + c^2

, the expression

\frac{{\sin 3B}}{{\sin B}}

simplifies to

{\left( {\frac{{{c^2} - {a^2}}}{{2ca}}} \right)^2}

More Questions from PROPERTIES OF TRIANGLES

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