If the base angles of a triangle are $22\frac{1^0}{2},112\frac{1^0}{2},$ then base and height are in the ratio

Let$B=22\frac{1^0}{2} C=112\frac{1^0}{2}$

$\therefore A={45}^0$

$\frac{a}{h}=\frac{a}{\left(\frac{2\Delta }{a}\right)}=\frac{a^2}{2\Delta }=\frac{\cancel{4}{R}^2{\sin }^{2}A}{\cancel{4}{R}^2\sin A\sin B\sin C}$

$=\frac{\sin A}{\sin B\text{ sinC}}$

$\begin{array}{l}=\frac{\sin {45}^0}{\sin 22\frac{1^0}{2}\text{ cos22}\frac{1^0}{2}}\\ =\frac{2\sin 22\frac{1^0}{2}\cos 22\frac{1^0}{2}}{\sin 22\frac{1^0}{2}\cos 22\frac{1^0}{2}}\\ =2\\ =\frac{2}{1}\\ a:h=2:1\end{array}$