Question

In a triangle ABC, a, b, A are given and $c_{1},c_{2}$ are two values of third side c. The sum of the areas of triangles with sides a, b, $c_{1}$ and $a,b,c_{2}$ is

• A. $\frac{1}{2}a^{2}\sin 2A$
• B. $\frac{1}{2}b^{2}\sin 2A$
• C. $b^{2}\sin 2A$
• D. $a^{2}\sin 2A$

Answer 1

Answer: (B)

$\frac{1}{2}b^{2}\sin 2A$

Answer 2

Let the triangles be $\Delta_{1} = ABC_{1}$ and $\Delta_{2} = ABC_{2}$ A, b, a are given and c has two values. Hence we apply cosine formulae $\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}$ or $c^{2} - 2bc\cos A + b^{2} - a^{2} = 0$.

Above is quadratic in c If $c_{1},c_{2}$ be the two values of c, then $c_{1} + c_{2} = 2b\cos A,c_{1}c_{2} = b^{2} - a^{2}$

$\Delta_{1} = \frac{1}{2}ab\sin C_{1}$, $\Delta_{2} = \frac{1}{2}ab\sin C_{2}$

$\therefore$ $\Delta_{1} + \Delta_{2} = \frac{1}{2}ab(\sin C_{1} + \sin C_{2}) = \frac{1}{2}abk(2b\cos A)$ $= b^{2}ak\cos A = b^{2}\sin A\cos A$ = $\frac{1}{2}b^{2}\sin 2A$.