# Find an angle between a triangle and a plane

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.

• Yes, that's what they're asking. However, i'm not supposed to use the formula for planes, or normal vectors for that matter. Jul 8, 2020 at 15:59
• $KD\perp AB$ by the Theorem of Three Perpendiculars. Jul 8, 2020 at 21:27
• @Aretino I think, methodical it's not so good theorem. It helps maybe for solving easy problems. Jul 9, 2020 at 3:46
• @MichaelRozenberg I'm afraid I don't understand what you wrote. I was just answering "How can I know for sure that 𝐾𝐷 is a perpendicular line?" Jul 9, 2020 at 6:47

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 $$\phi=30^{\circ}.$$

• Your answer is indeed correct. I'm having trouble understanding the 3rd step, could you explain the simplification? I'm aware that $AC \sin(\alpha) = h$ and $BC\sin(\beta) = h$ Jul 8, 2020 at 16:40
• So $AC\cdot BC$ would be $= \frac{h^2}{\sin(\alpha)\sin(\beta)}$ right? Jul 8, 2020 at 16:42
• Oh, you took the $h^2$ from the square root and divided numerator and denominator by h, thus we're left with no $h$. I understood now. Jul 8, 2020 at 16:46
• @Ebrin I saw you comment only now. Is it clear now? Jul 8, 2020 at 17:11
• @Ebrin Just if we want to prove that $a\perp b$ we need one of these lines to put in the plane and to prove that the second line is a perpendicular to the plane. We need to prove that $DK\perp AB$. We see that $DK\subset(DKC)$ and it's enough to prove that $AB\perp(DKC),$ which we made. Jul 9, 2020 at 7:11