5
$\begingroup$

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:

enter image description here

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.

$\endgroup$
4
  • $\begingroup$ 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 '20 at 15:59
  • 1
    $\begingroup$ $KD\perp AB$ by the Theorem of Three Perpendiculars. $\endgroup$ Jul 8 '20 at 21:27
  • $\begingroup$ @Aretino I think, methodical it's not so good theorem. It helps maybe for solving easy problems. $\endgroup$ Jul 9 '20 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 '20 at 6:47
2
$\begingroup$

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

$\endgroup$
6
  • $\begingroup$ 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 '20 at 16:40
  • $\begingroup$ So $AC\cdot BC $ would be $= \frac{h^2}{\sin(\alpha)\sin(\beta)}$ right? $\endgroup$
    – Ebrin
    Jul 8 '20 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 '20 at 16:46
  • $\begingroup$ @Ebrin I saw you comment only now. Is it clear now? $\endgroup$ Jul 8 '20 at 17:11
  • 1
    $\begingroup$ @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 '20 at 7:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.