Review question

# Can we find the lengths and angles in this diagram of a roof? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R6144

## Solution

We’ll drop the units in all calculations and keep in mind that all lengths are in feet.

In the diagram, which represents the roof of a house, $ABCD$ is a horizontal rectangle and $XY$ a line parallel to $AB$ and $DC$. $AB=40'$, $BC=XY=16'$ and $AX=BY=CY=DX=17'$. Calculate $(i)$ the height of $XY$ above the plane $ABCD$, …

The height we want is $FY$.

By using Pythagoras’ theorem on $BGY$, we have $GY= \sqrt{17^2-8^2}=15.$

$8, 15$ and $17$ is a Pythagorean triple. These often come up in questions about triangles and it can be really helpful to notice them.

Another example is $3, 4$ and $5$. Do you know any more?

In right-angled triangle $GFY$ we have $GF=12$ and the hypotenuse $GY=15$ so $GFY$ is similar to a triangle with sides $3,4,5$ and thus $FY=9.$

If we don’t spot this we could use Pythagoras for a second time.

$(ii)$ the angle of inclination to the horizontal of the face $AXYB$, …

The desired angle is $\angle XHE$. Looking at other lengths in the diagram, we can see that $\tan(\angle XHE) = \dfrac{9}{8}$, and so $\angle XHE = 48.3^\circ$.

… and $(iii)$ the angle of inclination to the horizontal of the edge $AX$.

The desired angle is $\angle XAE$. We can see that $\sin(\angle XAE) = \dfrac{9}{17}$, and so $\angle XAE = 32.0^\circ$.