Question

If $a,b,A$ be given in a triangle and $c_1$ and $c_2$ are two possible values of the third side such that $c_1^2+c_2^2+c_1{c}_2=a^2,$ then $A$ is equal to:

• A. ${30}^0$
• B. ${60}^0$
• C. ${90}^0$
• D. ${120}^0$

Answer 1

Answer: (B), (D)

The correct options are
B ${60}^0$
D ${120}^0$

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\Rightarrow c^2-2bc\cos A+b^2-a^2=0$

$\therefore c_1+c_2=2b\cos A,c_1{c}_2=b^2-a^2$

Given$:c_1^2+c_1{c}_2+c_2^2=a^2$

$\Rightarrow {(c_1+c_2)}^2-c_1{c}_2=a^2$

$\Rightarrow 4b^2{\cos }^{2}A-(b^2-a^2)=a^2$

$\Rightarrow 4b^2{\cos }^{2}A-b^2+a^2=a^2$

$\Rightarrow 4b^2{\cos }^{2}A=b^2$

$\Rightarrow 4{\cos }^{2}A=1$

$\Rightarrow 2(1+\cos 2A)=1$

$\Rightarrow 1+\cos 2A=\frac{1}{2}$

$\Rightarrow \cos 2A=\frac{1}{2}-1=\frac{-1}{2}=\cos \frac{2\pi }{3}$

$\therefore 2A=2n\pi \pm \frac{2\pi }{3}$

Hence $A=n\pi \pm \frac{\pi }{3}$ where $n\in I$

$\therefore A=\frac{\pi }{3}$ or $\frac{2\pi }{3}$

i.e., $A={60}^0$ or ${120}^0$