Review question

# Given a picture hanging from a wall, can we find these angles? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R5533

## Solution

A rectangular picture of width $\quantity{32}{cm}$ and height $\quantity{20}{cm}$ has its lower edge $AB$ horizontal against a vertical wall. It hangs at an angle of $15^\circ$ to the wall and may be taken to be of negligible thickness. It is supported by two equal strings $FD$, $FC$ of length $\quantity{20}{cm}$ attached to the upper corners $C$, $D$ of the picture and to a point $F$ on the wall vertically above the centre of $AB$.

Copy the given sketch, including the projections of $C$, $D$ and $E$, the mid-point of $CD$, onto the wall at $X$, $Y$ and $Z$ respectively.

Identify in terms of the letters the following angles and calculate them:

1. the angle between the two strings…

We know the length of two of the sides of the right triangle $DEF$.

The angle between $DF$ and $FE$ is half the angle $\alpha$ that we need, so we can use the definition of sine to give $\alpha = 2\sin^{-1} \frac{16}{20} \quad \left(=2\arcsin\frac{16}{20}\right) \quad \approx 106.3^\circ.$

1. the angle between the string $DF$ and the wall…

The angle between a line and a plane is given by the angle between the line and its perpendicular projection onto that plane, that is, the angle between the line $DF$ and $FY$.

So looking at the triangle $AYD$, we can use the definition of cosine to find the length of $YD$ as

$20 \cos 15^\circ \approx \quantity{5.2}{cm}.$

We can now obtain the angle $\beta$ from the triangle $FDY$ with help of the definition of the tangent, giving $\beta = \tan^{-1} (\cos 15^\circ) \quad (=\arctan(\cos15^\circ)) \quad \approx 14.5^\circ.$

1. the angle between the wall and the plane containing both strings.

Let’s call the angle between the plane of the strings and the wall $\gamma$, which is given by the angle between the lines $FZ$ and $FE$.

Looking at the triangle $EFZ$, we can calculate the length of $EF$ with help of Pythagoras as $\sqrt{20^2-16^2} = 12$.

We also know that $EZ$ has the same length as $DY$, which we have already calculated. Finally we use the definition of sine again to obtain $\gamma = \sin^{-1} \frac{20 \cos 15^\circ}{12} \quad \left(=\arcsin\frac{20 \cos 15^\circ}{12}\right) \quad \approx 25.6^\circ.$