Question

In $\Delta$$ABC$, if $a$, $b$ and $\angle A$ are given, then two triangles can be formed with third side $c_1$ and $c_2$. Find $|c_1-c_2|$

• A. $2\sqrt{a^2-b^2{\sin }^{2}A}$
• B. $2\sqrt{a^2+b^2{\sin }^{2}A}$
• C. $2\sqrt{a^2-b^2{\cos }^{2}A}$
• D. $2\sqrt{a^2+b^2{\cos }^{2}A}$

Answer 1

The correct option is A $2\sqrt{a^2-b^2{\sin }^{2}A}$

$cosA=\frac{b^2+c^2-a^2}{2bc}$
Rearranging the terms,
$c^2-2bccosA+b^2-a^2=0$
$c_1+c_2=2bcosA$
$c_1{c}_2=b^2-a^2$
$|c_1-c_2|=\sqrt{\left(\right(c_1+c_2{)}^2-4c_1{c}_2)}$
$=2\sqrt{b^{2}co{s}^{2}A-b^2+a^2}$
$=2\sqrt{a^2-b^{2}si{n}^{2}A}$
Hence, option A is correct.