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(A) Let the triangle be $\Delta ABC$ with base $BC = x$ and height $AD = h$. The angles at the base are $\angle B = 22.5^o$ and $\angle ACB = 112.5^o$

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(A) Let the triangle be $\Delta ABC$ with base $BC = x$ and height $AD = h$. The angles at the base are $\angle B = 22.5^o$ and $\angle ACB = 112.5^o$.Since $\angle ACB$ is an exterior angle to $\Delta ABD$,the angle $\angle ACD = 180^o - 112.5^o = 67.5^o$.In right-angled $\Delta ACD$,$\angle CAD = 90^o - 67.5^o = 22.5^o$.Thus,$\angle BAC = 180^o - (22.5^o + 112.5^o) = 45^o$.Using the sine rule in $\Delta ABC$,$\frac{x}{\sin 45^o} = \frac{AC}{\sin 22.5^o} \Rightarrow AC = \frac{x \sin 22.5^o}{\sin 45^o}$.In $\Delta ACD$,$h = AC \sin 67.5^o = \frac{x \sin 22.5^o \sin 67.5^o}{\sin 45^o}$.Using $\sin 67.5^o = \cos 22.5^o$,we get $h = \frac{x \sin 22.5^o \cos 22.5^o}{\sin 45^o} = \frac{x \sin 45^o}{2 \sin 45^o} = \frac{x}{2}$.Therefore,the ratio $\frac{h}{x} = \frac{1}{2}$.

If the two angles on the base of a triangle are $22.5^o$ and $112.5^o$,then the ratio of the height of the triangle to the length of the base is

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