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Q.
If in a triangle $A B C, a, b, c$ and angle $A$ are given and $c \sin A< a < c$, then
Trigonometric Functions
Solution:
From cosine rule, $\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}$
or $b^{2}-(2 c \cos A) \cdot b+\left(c^{2}-a^{2}\right)=0$
which is a quadratic equation in $b$.
Since, $c \sin A< a< c$ two triangles will be obtained, but this is possible when two values of third side are also obtained.
Clearly two values of side $b$ will be $b_{1}$ and $b_{2}$.
Let these be roots of above equation.
$\therefore$ Sum of roots $=b_{1}+b_{2}=2 c \cos A$