Formula to calculate a length to a point on hypotenuse according to given angle

Question

I have a right triangle:

• Height: y (value over 0)
• Width: y (value over 0)
• Angle: α (degrees, value between 0-90)

I need to find out the formula to count the length of x.

Answer 1

Yes, it's possible. The law of sines used on the lower triangle (the one spanned by $\alpha$) say that $$ \frac{\sin(135^\circ - \alpha)}{1} = \frac{\sin \alpha}{x} $$ where $135^\circ - \alpha$ is the measure of the top angle of the triangle (the lower left angle has measure $45^\circ$). So we have $$ x = \frac{\sin{\alpha}}{\sin(135^\circ - \alpha)} $$

Answer 2

Using picture above, if $AB=BC=y$ then using Pythagoras' formula $AC=y\sqrt2$ and $CD=y\sqrt2-x$. Hence, using sine rule on triangle $ABD$ we get \begin{align} \frac{x}{\sin\alpha}&=\frac{y}{\sin(135^\circ-\alpha)}\\ x&=\frac{y\sin\alpha}{\sin(135^\circ-\alpha)} \end{align} or using sine rule on triangle $BCD$ we get \begin{align} \frac{y\sqrt2-x}{\sin(90^\circ-\alpha)}&=\frac{y}{\sin(45^\circ+\alpha)}\\ y\sqrt2-x&=\frac{y\sin(90^\circ-\alpha)}{\sin(45^\circ+\alpha)}\\ x&=y\sqrt2-\frac{y\sin(90^\circ-\alpha)}{\sin(45^\circ+\alpha)}\\ &=y\sqrt2\left(1-\frac{\cos\alpha}{\sin\alpha+\cos\alpha}\right)\\ &=\frac{y\sqrt2\sin\alpha}{\sin\alpha+\cos\alpha} \end{align}