Question

In an isosceles triangle $ABC,AB=AC.$ If vertical angle $A$ is ${20}^0,$ then $a^3+b^3$ is equal to

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

Answer 1

Answer: (C)

$3c^{2}a$

$\because \angle A={20}^0$
$\therefore \angle B=\angle C={80}^0$
$\Rightarrow △ABC$ is isosceles.
$\Rightarrow AB=AC$
$\Rightarrow c=b$
Using sine rule,
$\therefore \frac{a}{\sin 20}=\frac{b}{\sin 80}=\frac{c}{\sin 80}$
$Rightarrow\frac{a}{\sin 20}=\frac{b}{\sin (90-10)}=\frac{c}{\sin 80}$
$\Rightarrow \frac{a}{\sin 20}=\frac{b}{\cos 10}=\frac{c}{\cos 10}$
$\Rightarrow a=\frac{b\sin 20}{\cos 10}=\frac{b.2\sin 10\cos 10}{\cos 10}=2b\sin {10}^0$
$\therefore a^3+b^3={\left(2b\sin 10\right)}^3+b^3$
$=8b^3{\sin }^{3}10+b^3$
$=b^3(8{\sin }^{3}10+1)$
$=b^3\left(2(4{\sin }^{3}10+1)\right)$
Using the trigonometric formula $\sin 3A=3\sin A-4{\sin }^{3}A$ or $4{\sin }^{3}A=3\sin A-\sin 3A$above we get
$=b^3(2(3\sin 10-\sin 30)+1)$
$=b^3(6\sin 10-2\times \frac{1}{2}+1)$
$=b^3\left(6\sin 10\right)$
$=3b^2\left(2b\sin 10\right)$
We know that $a=2b\sin 10$
$\therefore a^3+b^3=3b^2\left(2b\sin 10\right)$(from above)
$=3b^{2}a=3c^{2}a$ (as $b=c$)