Question

If in a $\Delta$ ABC $a,b$ and $\angle A$ are given and $C_1,C_2$ are the possible values of the third side, then

• A. $C_1+C_2=2bcosA$
• B. $C_1.C_2=(b^2-a^2)$
• C. $C_1^2+C_2^2-2C_1{C}_{2}cos2A=4a^{2}co{s}^{2}A$
• D. $C_1+C_2=2acosB$

Answer 1

Answer: (C)

$C_1^2+C_2^2-2C_1{C}_{2}cos2A=4a^{2}co{s}^{2}A$

$a,b$ and $\angle$ A are given in $\Delta$ ABC,
$\therefore \cos A=\frac{b^2+c^2-a^2}{2bc}$
$\Rightarrow c^2-2bccosA+(b^2-a^2)=0$
$\Rightarrow c^2-\left(2bcosA\right)c+(b^2-a^2)=0$
which is quadratic in $c$ and gives two values of $c$ and let these $c_1$ and $c_2$
$\Rightarrow c_1+c_2=2b\cos A,c_1{c}_2=b^2-a^2$
$\therefore c_1^2+c_2^2-2c_1{c}_2\cos 2A$
$=(c_1+c_2{)}^2-2c_1{c}_2(1+\cos 2A)$
$=4b^{2}co{s}^{2}A-2c_1{c}_2(1+\cos 2A)$
$=4b^{2}co{s}^{2}A-4b^{2}co{s}^{2}A+4a^2{\cos }^{2}A$
$=4a^2{\cos }^{2}A$