The hypotenuse $AB$ of triangle $ABC$ lies in plane $Q$. Sides $AC$ and $BC $, respectively, create angles $\alpha$ and $\beta$ towards the plane Q (meaning they are tilted towards the plane $Q$ with such angles). Find the angle between plane $Q$ and the plane of the triangle, given $\sin(\alpha) = \frac{1}{3} $ and $\sin(\beta)=\frac{\sqrt5}{6}$.
I'm really struggling with these kinds of problems and I can't seem to find any material in English that covers this topic. Only videos I found about planes use normal vectors and equation of the plane, which is not necessary for this.
The picture wasn't given but Here's my interpretation:
let $CK$ be the perpendicular line from point $C$ to plane $Q$. $CD$ is the height of triangle $ABC$. What I'm struggling to understand is what will the dihedral angle be in this case? Well, I know that the angle between two planes is the angle between two perpendicular lines of such planes. One of which must be $CD$, but what will the other line be? Is it $KD$? How can I know for sure that $KD$ is a perpendicular line?
Anyway, I don't think I'm understanding the problem clearly. If someone can provide a graphical solution, i'll be very thankful.
Best Answer
Because $AB\perp CD$ and $AB\perp CK$, which says $AB\perp(CDK)$ and from here $AB\perp DK$.
Let $CK=h$ and $\measuredangle CDK=\phi$.
Thus, $$\sin\phi=\frac{h}{DC}=\frac{h}{\frac{AC\cdot BC}{\sqrt{AC^2+BC^2}}}=\frac{1}{\frac{\frac{1}{\sin\alpha}\cdot\frac{1}{\sin\beta}}{\sqrt{\frac{1}{\sin^2\alpha}+\frac{1}{\sin^2\beta}}}}=\sqrt{\sin^2\alpha+\sin^2\beta}.$$ Can you end it now?
I got $\phi=30^{\circ}.$