Question

If the angles of a triangle are in the ratio $1:3:5$ and $\theta$denote the smallest angle, then the ratio of the largest side to the smallest side of the triangle is:

• A. $\frac{(\sqrt{3}\sin \theta +\cos \theta )}{2\sin \theta }$
• B. $\frac{(\sqrt{3}\cos \theta -\sin \theta )}{2\sin \theta }$
• C. $\frac{(\cos \theta +\sqrt{3}\sin \theta )}{2\sin \theta }$
• D. $\frac{(\sqrt{3}\cos \theta +\sin \theta )}{2\sin \theta }$

Answer 1

The correct option is D

$\frac{(\sqrt{3}\cos \theta +\sin \theta )}{2\sin \theta }$

Explanation for the correct option.

Step 1: Find the value of angles

Given that, the angles of a triangle are in the ratio $1:3:5$ & $\theta$ is the smallest angle.

Let angles be $A,B,C$.

Then, $A=x,B=3x,C=5x$

Now using the angle sum property of a triangle I.e; the sum of all angles in a triangle is ${180}^o$.

$$\begin{gathered} x+3x+5x={180}^o \\ 9x={180}^o \\ x={20}^o \end{gathered}$$

Therefore, $A={20}^o,B={60}^o,C={100}^o$

Step 2: Find the ratio of the largest side to the smallest side of the triangle

Now use the sine rule i.e; $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$.

Since here angle $C>\mathrm{angle}\mathrm{A}$ so we need to find $\frac{c}{a}$.

$\begin{array}{rcl}\frac{c}{a}& =& \frac{\sin C}{\sin A}\\ & =& \frac{\sin {100}^o}{\sin {20}^o}\\ & =& \frac{\sin \left({120}^o-\theta \right)}{\sin {20}^o}\\ & =& \frac{(\sin {120}^o\cos {20}^o-\cos {120}^o\sin {20}^o)}{\sin {20}^o}\\ & =& \frac{\left(\frac{\sqrt{3}}{2}\cos {20}^o+\frac{1}{2}\sin {20}^o\right)}{\sin {20}^o}\\ & =& \frac{\sqrt{3}\cos {20}^o+\sin {20}^o}{2\sin {20}^o}\\ & =& \frac{\sqrt{3}\cos \theta +\sin \theta }{2\sin \theta }\end{array}$

Hence, option D is correct.