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:

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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.

  • $\begingroup$ Yes, that's what they're asking. However, i'm not supposed to use the formula for planes, or normal vectors for that matter. $\endgroup$
    – Ebrin
    Jul 8, 2020 at 15:59
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    $\begingroup$ $KD\perp AB$ by the Theorem of Three Perpendiculars. $\endgroup$ Jul 8, 2020 at 21:27
  • $\begingroup$ @Aretino I think, methodical it's not so good theorem. It helps maybe for solving easy problems. $\endgroup$ Jul 9, 2020 at 3:46
  • $\begingroup$ @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?" $\endgroup$ Jul 9, 2020 at 6:47

1 Answer 1


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

  • $\begingroup$ 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 $ $\endgroup$
    – Ebrin
    Jul 8, 2020 at 16:40
  • $\begingroup$ So $AC\cdot BC $ would be $= \frac{h^2}{\sin(\alpha)\sin(\beta)}$ right? $\endgroup$
    – Ebrin
    Jul 8, 2020 at 16:42
  • $\begingroup$ 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. $\endgroup$
    – Ebrin
    Jul 8, 2020 at 16:46
  • $\begingroup$ @Ebrin I saw you comment only now. Is it clear now? $\endgroup$ Jul 8, 2020 at 17:11
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    $\begingroup$ @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. $\endgroup$ Jul 9, 2020 at 7:11

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