Question

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

• A. $3ab$
• B. $3a{b}^2$
• C. $3a^{2}b$
• D. $3$

Answer 1

The correct option is B $3a{b}^2$

Given vertical angle$={20}^o$

$\Rightarrow \angle B+\angle B+{20}^o=180\Rightarrow 2B=160$ [Isosceles triangle ]

$\Rightarrow \angle B=\angle C={80}^o$

Apply cosine formula at $B$ :-

$\cos {80}^o=\frac{a^2+b^2-b^2}{2ab}$

$\cos {80}^o=\frac{a}{2b}$

$\therefore \sin {10}^o=\frac{a}{2b}$ [ As $\cos {80}^o=\sin {10}^o$]

As we know that,

$\sin 3\theta =3\sin \theta -4{\sin }^3\theta$

Put $\theta ={10}^o$

$\sin {30}^o=3\sin {10}^o-4{\sin }^3{10}^o$

$\frac{1}{2}=3\times \frac{a}{2b}-4a\times \frac{a^3}{8b^3}$

$1=\frac{3a}{b}-\frac{a^3}{b^3}\Rightarrow 1=\frac{3a{b}^2-a^3}{b^3}$

$3a{b}^2-a^3=b^3$

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