In a right angled triangle $ABC$, where $C$ is the right angle, if $a = 50$ and $B = 75^{\circ}$, find the sides.
Given: $\;$ $\triangle ABC$, with $C = 90^{\circ}$, $B = 75^{\circ}$, $a = 50$
By cosine rule, $\;$ $\cos B = \dfrac{c^2 + a^2 - b^2}{2 c a}$; $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
Here,
$\cos B = \cos 75^{\circ} = \dfrac{c^2 + 50^2 - b^2}{2 \times 50 \times c}$
i.e. $\;$ $\dfrac{\sqrt{3} - 1}{2 \sqrt{2}} = \dfrac{c^2 + 2500 - b^2}{2 \times 50 \times c}$
i.e. $\;$ $\dfrac{\sqrt{3} - 1}{\sqrt{2}} = \dfrac{c^2 + 2500 - b^2}{50 \times c}$ $\;\;\; \cdots \; (1)$
and,
$\cos C = \cos 90^{\circ} = \dfrac{50^2 + b^2 - c^2}{2 \times 50 \times b}$
i.e. $\;$ $2500 + b^2 - c^2 = 0$
i.e. $\;$ $c^2 - b^2 = 2500$ $\;\;\; \cdots \; (2)$
$\therefore \;$ In view of equation $(2)$, equation $(1)$ becomes
$\dfrac{\sqrt{3} - 1}{\sqrt{2}} = \dfrac{2500 + 2500}{50 c}$
i.e. $\;$ $\dfrac{\sqrt{3} - 1}{\sqrt{2}} = \dfrac{5000}{50 c}$
i.e. $\;$ $\dfrac{\sqrt{3} - 1}{\sqrt{2}} = \dfrac{100}{c}$
$\implies$ $c = \dfrac{100 \sqrt{2}}{\sqrt{3} - 1} = 193.15$
Substituting the value of $c$ in equation $(2)$ gives
$b^2 = c^2 - 2500 = 193.15^2 - 2500 = 34806.92$
$\implies$ $b = \sqrt{34806.92} = 186.57$
$\therefore \;$ The sides of the triangle are $\;$ $a = 50$, $b = 186.57$, $c = 193.15$