question_answer

If in an isosceles triangle with base 'a' vertical angle \[20{}^\circ \]and lateral side each of length 'b' is given, then the value of \[{{a}^{3}}+{{b}^{3}}\]equals

A) \[3ab\]  

B) \[3a{{b}^{2}}\]

C) \[3{{a}^{2}}b\]

D) \[3\]

Correct Answer: B

Solution :

In \[\Delta ABD,\]

\[\sin 10{}^\circ =\frac{BD}{AB}=\frac{9a}{2b}\]

\[\sin 30{}^\circ =3\sin 10{}^\circ -4{{\sin }^{3}}10{}^\circ \]

\[\Rightarrow \]   \[\frac{1}{2}=\frac{3a}{2b}-\frac{4{{a}^{3}}}{8{{b}^{3}}}\]

\[\Rightarrow \]   \[{{b}^{3}}=3a{{b}^{2}}-{{a}^{3}}\]

\[\Rightarrow \]   \[{{a}^{3}}+{{b}^{3}}=3a{{b}^{2}}\]