Question

If in a triangle $ABC,a,b,c$ and angle $A$ are given and $c\sin A<a<c$, then

• A. $b_1+b_2=2c\cos A$
• B. $b_1+b_2=c\cos A$
• C. $b_1+b_2=3c\cos A$
• D. $b_1+b_2=4c\sin A$

Answer 1

Answer: (A)

From cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2bc}$
or $b^2-(2c\cos A)\cdot b+\left(c^2-a^2\right)=0$
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 $=b_1+b_2=2c\cos A$