Question

If two angles of a $△ABC$ are ${45}^o$ and ${60}^o$, then the ratio of the smallest and the greatest sides are

• A. $(\sqrt{3}-1):1$
• B. $\sqrt{3}:\sqrt{2}$
• C. $1:\sqrt{3}$
• D. $\sqrt{3}:1$

Answer 1

The correct option is A $(\sqrt{3}-1):1$

$\to$ Given two angles of a triangle are ${45}^o$ and ${60}^o$.

$\therefore$ The third angle is ${180}^o-{105}^o={75}^o$.

We know that side opposite to the smallest angle is the smallest side and side opposite to the largest angle is the largest side.

we know from sine rule in a triangle.

$\frac{c}{\sin {75}^o}=\frac{a}{\sin {45}^o}$

$\because (in△ABC\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C})$

we know that $\sin {75}^o=\frac{\sqrt{3+1}}{2\sqrt{2}}$.

$\therefore \frac{\left(c\right)\left(2\sqrt{2}\right)}{\sqrt{3+1}}=a\sqrt{2}\Rightarrow \frac{a}{c}=\frac{2}{\sqrt{3}+1}$

by rationalizing the denominator

$\therefore \frac{a}{c}=\frac{\left(2\right)(\sqrt{3}-1)}{2}$

$\therefore a:c=\sqrt{3}-1:1$.