Right triangle ABC has hypotenuse AC, angle CAB=30°, and BC=√2. Right triangle ACD has hypotenuse AD and angle DAC=45°. The interiors of ABC and ACD do not overlap. Find the length of the perpendicular from D onto AB.
Can someone please provide a solution to this problem? Any help is greatly appreciated.
We have $$sin 30^o = \frac{1}{2} = \frac{BC}{AC} = \frac{\sqrt{2}}{AC}$$
so $AC = 2\sqrt2$.
Similarly $$cos 45^o = \frac{1}{\sqrt2} = \frac{AC}{AD} = \frac{2\sqrt2}{AD}$$
so $AD = 4$.
Finally, let $D_p$ denote described perpendicular. Then $$ sin 75^o = \frac{D_p}{AD}$$
so $D_p = 4 \times sin 75^o$.