Question

In ambiguous case if a, b and A are given and if there are two possible values of third side, are c1 and c2, then

• A. $c_1-c_2=2\sqrt{(a^2-b^{2}si{n}^{2}A)}$
• B. $c_1-c_2=3\sqrt{(a^2-b^{2}si{n}^{2}A)}$
• C. $c_1-c_2=4\sqrt{(a^2+b^{2}si{n}^{2}A)}$
• D. $c_1-c_2=2\sqrt{(a^2+b^{2}si{n}^{2}A)}$

Answer 1

The correct option is A $c_1-c_2=2\sqrt{(a^2-b^{2}si{n}^{2}A)}$

We know that $cosA=\frac{b^2+c^2-a^2}{2bc}$

$⟹c^2-2b\left(cosA\right)c+(b^2-a^2)=0$ -----(1)

eqn (1) is a quadratic equation. Let the roots of the equation be $c_1,c_2$

Sum of the roots is $c_1+c_2=2bcosA$

and product of the roots is $c_1{c}_2=b^2-a^2$

Therefore, $c_1-c_2=\sqrt{(c_1+c_2{)}^2-4c_1{c}_2}$

$=\sqrt{(2bcosA{)}^2-4(b^2-a^2)}$

$=\sqrt{(4a^2-4b^2(1-co{s}^{2}A\left)\right)}$

$=2\sqrt{(a^2-b^{2}si{n}^{2}A)}$