In an isosceles $\Delta \;ABC,AB = AC$. If vertical angle $A$ is $20^\circ $, then ${a^3} + {b^3}$ is equal to

JEE MATHEMATICS MCQ: PROPERTIES OF TRIANGLES

1 Easy

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Analyzing +4 | -1

In an isosceles

\Delta \;ABC,AB = AC

. If vertical angle

A

20^\circ

, then

{a^3} + {b^3}

is equal to

View answer and explanation

Correct option: C

3{c^2}a

Option C (

3{c^2}a

) is correct because using the sine rule and trigonometric identities, it can be shown that

{a^3} + {b^3} = 3a{c^2}

when AB=AC and angle A is 20 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