The sides of a triangle are $9$ and $3$, and the difference of the angles opposite to them is $90^\circ$. Find the base and the angles.
Let the given sides of the triangle be $a = 9$ and $b = 3$.
Difference of angles opposite to sides $a$ and $b$ is $A - B = 90^\circ$
Now, $\;$ $\tan \left(\dfrac{A - B}{2}\right) = \left(\dfrac{a - b}{a + b}\right) \cot \left(\dfrac{C}{2}\right)$
i.e. $\;$ $\tan \left(\dfrac{90^\circ}{2}\right) = \left(\dfrac{9 - 3}{9 + 3}\right) \cot \left(\dfrac{C}{2}\right)$
i.e. $\;$ $\cot \left(\dfrac{C}{2}\right) = \dfrac{12 \times \tan \left(45^\circ\right)}{6} = 2$
i.e. $\;$ $\dfrac{C}{2} = \cot^{-1} \left(2\right) = 26.5651^\circ$
$\implies$ $C = 53.1302^\circ = 53^\circ 7' 48.72''$
By cosine rule, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$
i.e. $\;$ $c^2 = a^2 + b^2 - 2 a b \cos C$
i.e. $\;$ $c^2 = 9^2 + 3^2 - 2 \times 9 \times 3 \times \cos \left(53^\circ 7' 48.72''\right)$
i.e. $\;$ $c^2 = 57.6054$ $\implies$ $c = 7.5898$
In $\triangle ABC$, $\;$ $A + B = 180^\circ - C$
i.e. $A + B = 180^\circ - 53^\circ 7' 48.72'' = 126^\circ 52' 11''$ $\;\;\; \cdots \; (1a)$
Given: $\;$ $A - B = 90^\circ$ $\;\;\; \cdots \; (1b)$
Adding equations $(1a)$ and $(1b)$ we get,
$2 A = 216^\circ 52' 11''$ $\implies$ $A = 108^\circ 26' 6''$
Substituting the value of $A$ in equation $(1b)$ gives
$B = A - 90^\circ = 108^\circ 26' 6'' - 90^\circ = 18^\circ 26' 6''$
$\therefore \;$ $A = 108^\circ 26' 6''$, $\;$ $B = 18^\circ 26' 6''$, $\;$ $C = 53^\circ 7' 48.72''$, $\;$ $c = 7.5898$