Question

If in a $∆\mathrm{ABC}$, $4\mathrm{sinA}=4\mathrm{sinB}=3\mathrm{sinC}$, then $\mathrm{cosC}$ is equal to

• A. $\frac{1}{3}$
• B. $\frac{1}{9}$
• C. $\frac{1}{27}$
• D. $\frac{1}{8}$

Answer 1

The correct option is B

$\frac{1}{9}$

Explanation for the correct option.

Find the value of $\mathrm{cosC}$:

Given,

$4\sin A=4\sin A=3\sin C$.

Let $4\sin A=4\sin B=3\sin C=k$

$\Rightarrow \sin A=\frac{k}{4},\sin B=\frac{k}{4}$and $\sin C=\frac{k}{3}$.

Now use sine rule,

$$\begin{gathered} \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=\lambda \\ \Rightarrow a=\lambda \sin A,b=\lambda \sin B,c=\lambda \sin C \end{gathered}$$

Substituting the values of $\sin A,\sin B$and $\sin C$.

$\Rightarrow a=\lambda \frac{k}{4},b=\lambda \frac{k}{4},c=\lambda \frac{k}{3}$

Let, $\lambda k=p$.

$\Rightarrow a=\frac{p}{4},b=\frac{p}{4},c=\frac{p}{3}$

Now, we know that,

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

Substituting the values of $a,b$ and $c$.

$\begin{array}{rcl}\cos C& =& \frac{{\left({\displaystyle \frac{p}{4}}\right)}^2+{\left({\displaystyle \frac{p}{4}}\right)}^2-{\left({\displaystyle \frac{p}{3}}\right)}^2}{2\left({\displaystyle \frac{p}{4}}\right)\left({\displaystyle \frac{p}{4}}\right)}\\ & =& \frac{{\displaystyle \frac{p^2}{16}}+{\displaystyle \frac{p^2}{16}}-{\displaystyle \frac{p^2}{9}}}{{\displaystyle \frac{p^2}{8}}}\\ & =& \frac{{\displaystyle \frac{1}{16}+\frac{1}{16}-\frac{1}{9}}}{{\displaystyle \frac{1}{8}}}\\ & =& \frac{1}{9}\end{array}$

Hence, the correct option is B.