Question

In $△ABC,$ if $\angle A={45}^o$ , $\angle B={60}^o$ , then find the ratio of its sides.

Answer 1

$\angle A={45}^o,\angle B={60}^o$

By Angle sum property of a triangle

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

${45}^o+{60}^o+\angle C={180}^o$

$\Rightarrow \angle C={180}^o-{105}^o={75}^o$

By sine rule, $a:b:c=\sin A:\sin B:\sin C$

$\Rightarrow a:b:c=\sin {45}^o:\sin {60}^o:\sin {75}^o$

$\Rightarrow a:b:c=\frac{1}{\sqrt{2}}:\frac{\sqrt{3}}{2}:\sin (90-{15}^o)$

$\Rightarrow a:b:c=\frac{1}{\sqrt{2}}:\frac{\sqrt{3}}{2}:\cos {15}^o$

$=\frac{2}{2\sqrt{2}}:\frac{\sqrt{3}}{2\sqrt{3}}:\frac{\sqrt{3}-1}{2\sqrt{2}}$

$\therefore a:b:c=2:\sqrt{6}:\sqrt{3}-1$

(by making the denominators $2\sqrt{2}$).