Q24 / 50 · Manipal (MET) 2024 · Mathematics
In triangle $A B C, \sin A, \sin B$ and $\sin C$ are in $A P$ $\left(C\gt90^{\circ}\right)$, then $\cos A$ is
- A$\frac{3 c-4 b}{2 b}$
- B$\frac{3 c-4 b}{2 c}$
- C$\frac{4 c-3 b}{2 b}$
- Correct answerD$\frac{4 c-3 b}{2 c}$
Solution
As we know,
$\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$
and $\sin A, \sin B$ and $\sin C$ are in $A P$.
So, $a, b, c$ will also be in AP
$\begin{aligned}
\therefore 2 b & =a+c \\
\text { and } \cos A & =\frac{b^2+c^2-a^2}{2 b c}=\frac{b^2+c^2-(2 b-c)^2}{2 b c} \\
& =\frac{b^2+c^2-\left(4 b^2+c^2-4 b c\right)}{2 b c} \\
& =\frac{-3 b^2+4 b c}{2 b c} \\
& =\frac{b(4 c-3 b)}{2 b c}=\frac{4 c-3 b}{2 c}
\end{aligned}$
Hence, the correct answer is option: 'D'.