In $\Delta \;ABC,{\rm{if}}\left( {a + b + c} \right)\left( {a - b + c} \right) = 3ac$, then

$\angle A + \angle C = 90^\circ $

JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

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Application +4 | -1

\Delta \;ABC,\mathrm{if}\left( {a + b + c} \right)\left( {a - b + c} \right) = 3ac

, then

View answer and explanation

Correct option: A

\angle B = 60^\circ

Option A (

\angle B = 60^\circ

) is correct because by simplifying the given equation and using the cosine rule, we find that angle B is 60 degrees.

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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}

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

\Delta

\Delta \;ABC

\tan A

A

2 s