Question

If two angles of $∆\mathrm{ABC}$ are ${45}^{°}$ and ${60}^{°}$, then the ratio of the smallest and the greatest sides are

Answer 1

Finding the ratio of smallest and greatest sides:

According to the property of a triangle greatest side of a triangle is opposite to its greatest angle.

And also, the smallest side of a triangle is opposite to its smallest angle.

Given two angles $45°$, $60°$ then the third angle is

$\left(180°-\left(45°+60°\right)\right)=75°$

Smallest angle is ${45}^{°}$ and the largest angle is ${75}^{°}$

By sine rule

$\frac{\mathbf{a}}{\mathbf{sinA}}\mathbf{=}\frac{\mathbf{b}}{\mathbf{sinB}}\mathbf{=}\frac{\mathbf{c}}{\mathbf{sinC}}$

$\therefore$Required ratio $=\frac{\sin {45}^{°}}{\sin {75}^{°}}$

$\begin{array}{rcl}& =& \frac{\frac{1}{\sqrt{2}}}{{\displaystyle \frac{\sqrt{3}+1}{2\sqrt{2}}}}\left[\because \sin 75°=\frac{\sqrt{3}+1}{2\sqrt{2}}\right]\\ & =& \frac{2}{\sqrt{3}+1}\end{array}$

$\begin{array}{rcl}& =& \frac{2}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}\mathbf{}\left[\mathrm{by}\mathrm{rationalizing}\right]\\ & =& \frac{2\left(\sqrt{3}-1\right)}{{\left(\sqrt{3}\right)}^2-1}\mathbf{[}\mathbf{\because }\mathbf{}\left(a+b\right)\left(a-b\right)\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{-}{\mathbf{b}}^{\mathbf{2}}\mathbf{]}\\ & =& \frac{2\left(\sqrt{3}-1\right)}{3-1}\\ & =& \sqrt{3}-1\end{array}$

Hence, the ratio of the smallest and the greatest sides is $\sqrt{\mathbf{3}}\mathbf{-}\mathbf{1}$.