I've been working through a trigonometry book and have been stuck on the following question for a while now.

The diagram shows a roof structure. PQRS is a horizontal rectangle. The faces ABRQ, ABSP, APQ and BRS all make an angle of $45$ degrees with the horizontal. Find the angle made by the sloping edges with the horizontal*

Roof Structure Image

The answer they're looking for is $35.26$ degrees.

I believe that this angle is $AQC$ where $C$ is the point directly below $A$. In order to find this the hypotenuse length $AQ$ is needed and either the height $CA$ or width $QC$.

I've started by finding the sloping lengths of the middle of each face, because there's a $45$ degree angle the height/width will be the same so I tried substituting in $1$ resulting in a sloping length of $2$.

Using the width / face slope I tried calculating $AQ$ by creating a new triangle halving the $PAQ$ face with a height of $2$ and width of $1$ which gives $AQ$ as $2.2$. The length $QC$ comes out as $2$ using Pythagoras.

These values for the triangle $ACQ$ give me an angle of $24.6$ degrees which is incorrect.

I think either my idea of substitution is wrong or the way I'm breaking up the structure.

Any ideas / hints would be greatly appreciated, thank you.

Link to question here with diagram if needed(Section 3.1 Exercise 10).

  • $\begingroup$ If a rooster lays an egg ... $\endgroup$ – John Joy Jan 7 '16 at 16:50

Let's try to simplify this problem a little bit. Think of a square pyramid with side length 2. If the angle between the sloped planes and the plane of the base is 45 degrees for all the sloped faces, then the height of the pyramid would be 1. (To see this, draw a right isosceles triangle with the hypotenuse the vertical median of one of the sloped faces.) Next, we find the length of the slanted edge to be $\sqrt{3}$ by Pythagoras. We then observe that the distance from the center of the base to a vertex of the base is $\sqrt{2}$. We now have that $\tan(x) = \frac{1}{\sqrt{2}}$, so $x = \arctan(\frac{1}{\sqrt{2}})$, which is about 35.264 degrees. Hope this helped!

  • $\begingroup$ Thanks, breaking it down into a pyramid makes things much easier - Looks like I was forgetting to take the square root when finding the distance from vertex to center. $\endgroup$ – guest501051 Jan 8 '16 at 1:01

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