Question

If, $\sin A:\sin B:\sin C=3:4:5$, then $\cos A:\cos B$ is equal to

• A. $4:3$
• B. $5:3$
• C. $3:4$
• D. $3:5$

Answer 1

The correct option is A

$4:3$

Explanation for the correct option.

Step 1: Find the length of the sides

Given that, $\sin A:\sin B:\sin C=3:4:5$

Let, $\sin A=3x,\sin B=4x,\sin C=5x$

Using sine rule i.e; $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ we have:

$\frac{a}{3x}=\frac{b}{4x}=\frac{c}{5x}=k\left(\mathrm{constant}\right)$

$\Rightarrow a=3xk,b=4xk,c=5xk$

Step 2: Find $\cos A:\cos B$

To find this we use the cosine rule i.e; $\cos A=\frac{\left(b^2+c^2-a^2\right)}{2bc}$.

$\begin{array}{rcl}\cos A& =& \frac{{\left(4xk\right)}^2+{\left(5xk\right)}^2-{\left(3xk\right)}^2}{2\left(4xk\right)\left(5xk\right)}\\ & =& \frac{16x^2{k}^2+25x^2{k}^2-9x^2{k}^2}{40x^2{k}^2}\\ & =& \frac{32x^2{k}^2}{40x^2{k}^2}\\ & =& \frac{4}{5}\end{array}$

Similarly for $\cos B$.

$\begin{array}{rcl}\cos B& =& \frac{{\left(3xk\right)}^2+{\left(5xk\right)}^2-{\left(4xk\right)}^2}{2\left(3xk\right)\left(5xk\right)}\\ & =& \frac{15x^2{k}^2+25x^2{k}^2-16x^2{k}^2}{30x^2{k}^2}\\ & =& \frac{18x^2{k}^2}{40x^2{k}^2}\\ & =& \frac{3}{5}\end{array}$

So, $\begin{array}{rcl}\cos A:\cos B& =& \frac{4}{5}:\frac{3}{5}\\ & =& 4:3\end{array}$

Hence, option A is correct.