Question:
If in ∆ABC, ∠A = 45°, ∠B = 60° and ∠C = 75°, find the ratio of its sides.
Solution:
Let $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$
Then,
$\frac{a}{\sin 45^{\circ}}=\frac{b}{\sin 60^{\circ}}=\frac{c}{\sin 75^{\circ}}=k$
$\Rightarrow \frac{a}{\frac{1}{\sqrt{2}}}=\frac{b}{\frac{\sqrt{3}}{2}}=\frac{c}{\frac{1}{2 \sqrt{2}}(1+\sqrt{3})} \quad\left[\because \sin 75^{\circ}=\sin \left(30^{\circ}+45^{\circ}\right)=\sin 30^{\circ} \cos 45^{\circ}+\sin 45^{\circ} \cos 30^{\circ}\right]$
On multiplying by $2 \sqrt{2}$, we get:
$a: b: c=2: \sqrt{6}:(1+\sqrt{3})$
Hence, the ratio of the sides is $2: \sqrt{6}:(1+\sqrt{3})$.