Height and Distance: Angle Tricks

If the angles from two points in line with the tower are complementary, the height is the geometric mean of the two distances.

The idea

On a line through the foot of a tower there are two points at distances $a$ and $b$ from the foot, and the angles of elevation of the top from them are complementary (they add to $90^{\circ }$). If the angle at the far point is $\theta$, the angle at the near point is $90^{\circ }-\theta$. From the far point $\tan \theta =\frac{h}{b}$ and from the near point $\tan (90^{\circ }-\theta )=\cot \theta =\frac{h}{a}$. Their product is $\tan \theta \cdot \cot \theta =1$, that is $\frac{h}{b}\times \frac{h}{a}=1$ and $h^2=ab$. No $\theta$ and no value of $\tan$ is needed.

The trick works even if the points are on opposite sides, because $\tan$ only sees the distance, not the side.

Steps

1. Check that the two angles add to $90^{\circ }$.
2. Take both distances $a$ and $b$ from the foot, not between the points.
3. Write $h=\sqrt{ab}$. For a missing distance use $b=\frac{h^2}{a}$.

If the angle doubles on coming nearer, the slant distance to the top equals the distance walked.

The idea

From the far point $A$ the angle to the top $T$ is $\alpha$, and from the nearer point $B$ it is $\beta$. Look at triangle $ABT$. The angle $\beta$ at $B$ is an exterior angle of this triangle (extend $AB$ beyond $B$ and you get the line of the ground). An exterior angle equals the sum of the two far interior angles: $\beta =\alpha +\angle ATB$, so $\angle ATB=\beta -\alpha$. Now if $\beta =2\alpha$, then $\angle ATB=\alpha$. The triangle has two equal angles ($\alpha$ at $A$ and $\alpha$ at $T$), so the sides opposite them are equal: $BT=AB$.

Here $d$ is the distance walked ($AB$); the distance from the near point to the foot $c=d\cos \beta$ also comes from it. It works even if the angle $\alpha$ is unknown: $h=\sqrt{d^2-c^2}$ (Pythagoras).

Steps

1. Check that the near angle is exactly twice the far angle.
2. Then write $BT=AB=d$: this is the hypotenuse of the right triangle at the near point.
3. Then $h=d\sin \beta$. Or, if $\alpha$ is unknown, take the distance $c$ from the near point to the foot and use $h=\sqrt{d^2-c^2}$.

When a fraction of $\tan$ or $\sin$ is given instead of an angle: take the height as an LCM of units and read off the bases.

The idea

Sometimes the angle is not given, only its $\tan$ as a fraction: $\tan \alpha =\frac{3}{8}$, $\tan \beta =\frac{2}{3}$. The fraction means height : base $=3:8$. Both triangles have the same height but different numerators (3 and 2), so the bases cannot be added or subtracted directly. The cure: take the height as the LCM of the numerators ($\text{LCM}(3,2)=6$). Then the height is 6 units in both triangles, and the bases come out as $6\times \frac{8}{3}=16$ and $6\times \frac{3}{2}=9$ units.

Here: “opposite : adjacent” is the ratio of the side opposite the angle (the height) to the side adjacent to it (the base); if $\sin$ or $\cos$ is given, first turn it into $\tan$.

Steps

1. Write each $\tan$ as “opposite : adjacent” (if $\sin /\cos$ is given, make the $\tan$ from the triple).
2. Take the height = LCM of the numerators (in units) and find both bases.
3. Subtract the bases for the same side, add for opposite sides, and equate to the given distance.
4. Find the value of one unit; height $=$ LCM $\times$ unit.

When two parts (a building and the flag on it) subtend equal angles at the eye: the double-angle formula.

The idea

Suppose from a ground point $O$ a building (height $b$) and the flagstaff (length $f$) on it subtend equal angles. The building makes the angle $\alpha$ at $O$ (foot to top) and the flag the same $\alpha$ above it. Then the angle of elevation of the whole ($b+f$) is $2\alpha$. So $\tan \alpha =\frac{b}{x}$ and $\tan 2\alpha =\frac{b+f}{x}$, where $x$ is the distance to the foot. Putting in $\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }$ brings everything into $x$.

Instead of memorising the formula it is easier to make a fraction of $\tan \alpha =\frac{b}{x}$ and get $\tan 2\alpha$. A few matches to remember: $\tan \alpha =\frac{1}{3}\Rightarrow \tan 2\alpha =\frac{3}{4}$ and $b:f=4:5$; $\tan \alpha =\frac{1}{2}\Rightarrow \tan 2\alpha =\frac{4}{3}$ and $b:f=3:5$.

Steps

1. Write the angles of the building and of the whole height as $\alpha$ and $2\alpha$.
2. Form $\tan \alpha =\frac{b}{x}$ and $\tan 2\alpha =\frac{b+f}{x}$.
3. Put in $\tan 2\alpha =\frac{2\tan \alpha }{1-{\tan }^2\alpha }$ and find $x^2$.

Looking at two vertical things from a point between them, or from the top of one building at the other.

Looking at a tower from the foot and from the top of a tree; raising an incomplete pillar.

Straight slanted lines: a ladder, a wire, a railing. Height $l\sin \theta$ and distance $l\cos \theta$.

A tree broken by a storm: the standing part and the bent part add up to the total height.

Height and distance along a sloping path, and a peak whose angle changes while you climb.

A cloud (or bird, star) and its reflection in the water: as high above as the image is below.

A sphere subtending $\alpha$ at the eye with its centre at an angle of elevation $\beta$: the height of the centre.

Every method: how to recognise the question and which formula to use.

Exam-style mixed questions: first recognise the trick or method.

The most important points; tap to see the answer.