Question

In a triangle $ABC$, $\sin A:\sin B:\sin C=1:2:3$. If $b=4cm$, then the perimeter of the triangle is

• A. $6cm$
• B. $24cm$
• C. $12cm$
• D. $8cm$

Answer 1

The correct option is C

$12cm$

Explanation for the correct option:

Step 1: Obtain the equations.

It is given that, $\sin A:\sin B:\sin C=1:2:3$

Write $\sin B$ in terms of $\sin A$.

$$\begin{gathered} \sin A:\sin B=1:2 \\ \Rightarrow \sin B=2\sin A...1 \end{gathered}$$

Write $\sin B$ in terms of $\sin A$.

$$\begin{gathered} \sin B:\sin C=2:3 \\ \Rightarrow \sin B=\frac{2}{3}\sin C...2 \end{gathered}$$

Step 2: Find the required value.

According to the sine rule:

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

Therefore,

$\frac{a}{\sin A}=\frac{b}{\sin B}...3$

From equation $1$ and equation $3$.

$$\begin{gathered} \Rightarrow \frac{a}{\sin A}=\frac{4}{2\sin A} \\ \Rightarrow a=2 \end{gathered}$$ { since, $b=4cm$}

So, the value of $a$ is $2cm$

Similarly,

$\frac{c}{\sin C}=\frac{b}{\sin B}...4$

From equation $2$ and equation $4$.

$$\begin{gathered} \Rightarrow \frac{c}{\sin C}=\frac{4}{{\displaystyle \frac{2}{3}\sin C}} \\ \Rightarrow c=6 \end{gathered}$$ { since, $b=4cm$}

So, the value of $c$ is $6cm$

The perimeter $P$ of the triangle is:

$$\begin{gathered} P=a+b+c \\ \Rightarrow P=12 \end{gathered}$$

Therefore, the perimeter of the triangle is $12cm$.