Question

In an ambiguous case of solving a triangle when $a=\sqrt{5},b=2,\angle A=\frac{\pi }{6}$ and the two possible values of third side are $c_1$ and $c_2$ then

• A. $|c_1-c_2|=2\sqrt{6}$
• B. $|c_1-c_2|=4\sqrt{6}$
• C. $|c_1-c_2|=4$
• D. $|c_1-c_2|=6$

Answer 1

Answer: (C)

$|c_1-c_2|=4$

$\because \cos A=\frac{b^2+c^2-a^2}{2bc}$
$\Rightarrow c^2-2bc\cos A+(b^2-a^2)=0$ .................$\left(1\right)$
Let $c_1$ and $c_2$ be the roots of eqn$\left(1\right)$
$\Rightarrow c_1+c_2=2b\cos A=2\times 2\times \cos {60}^0=2\times 2\times \frac{\sqrt{3}}{2}=2\sqrt{3}$ (Given $\angle A=\frac{\pi }{6}$)
and $c_1{c}_2=b^2-a^2=4-5=-1$ (Given:$a=\sqrt{5},b=2,$)
$\therefore |c_1-c_2|=\sqrt{{(c_1+c_2)}^2-4c_1{c}_2}$
$=\sqrt{12+4}=\sqrt{16}=4$