Question

The angles of a triangle are in the ratio $3:5:10$. Then, the ratio of the smallest side to the greatest side is

• A. $1:\sin {10}^{°}$
• B. $1:2\sin {10}^{°}$
• C. $1:\cos {10}^{°}$
• D. $1:2\cos {10}^{°}$

Answer 1

The correct option is D

$1:2\cos {10}^{°}$

Explanation for the correct option:

Finding the the ratio of the smallest side to the greatest side:

In $∆ABC$

$\Rightarrow \angle A+\angle B+\angle C={180}^{°}$

Let the angle ratio be $3x,5x,10x$

$$\begin{gathered} \Rightarrow 3x+5x+10x={180}^{°} \\ \Rightarrow 18x=180° \\ \Rightarrow x=10° \end{gathered}$$

Using sine rule

$$\begin{gathered} \Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k \\ \Rightarrow \frac{a}{c}=\frac{k\sin A}{k\sin C} \\ \Rightarrow \frac{a}{c}=\frac{\sin 3x}{\sin 10x}\left[\because A=3x;C=10x\right] \\ \Rightarrow \frac{a}{c}=\frac{\sin 3\times 10°}{\sin 2\times 10°}\left[\because x=10°\right] \\ \Rightarrow \frac{a}{c}=\frac{\sin {30}^{°}}{\sin {100}^{°}} \\ =\frac{{\displaystyle \frac{1}{2}}}{\sin (90°+10°)}\left[\because sin{30}^{°}=\frac{1}{2}\right] \\ \Rightarrow \frac{a}{c}=\frac{1}{2\cos 10°}\left[\because sin(90°+\theta )=cos\theta \right] \\ \therefore a:c=1:2\cos 10 \end{gathered}$$

Hence, option (D) is the correct answer.