Mixed configurations

The final family of heights-and-distances problems mixes the building blocks of the previous topics: an observer on a building looking at a tower, two towers with an observer between them, a hill with a flagstaff and a slope, and so on. These problems test whether you can identify the correct triangle network , not one triangle, but two or three sharing sides.

Once you locate the triangles and label common sides consistently, the algebra is no different from before. The trick is in the picture-making.

Two observers, one object

A common configuration: a building of height $b$ has an observer at its top; a tower of height $h$ stands $d$ metres away (on level ground). The observer on the building sees the top of the tower at angle of elevation $\alpha$ and its foot at angle of depression $\beta$ . Find $h$ and $d$ .

From the geometry (drawing the right triangles carefully):

$\tan\beta = b/d \Rightarrow d = b/\tan\beta = b\cot\beta$ .
$\tan\alpha = (h-b)/d \Rightarrow h = b + d\tan\alpha = b + b\cot\beta \tan\alpha$ .

So $h = b(1 + \cot\beta \tan\alpha) = b(\tan\beta + \tan\alpha)/\tan\beta$ .

One observer, two objects on opposite sides

A tower is in the middle of a road. From its top, the depressions of two cars on opposite sides of the tower are $\alpha$ and $\beta$ . The tower is $H$ tall. Distances of the two cars from the foot are

$x_A = H\cot\alpha, \quad x_B = H\cot\beta.$

The cars are $x_A + x_B = H(\cot\alpha + \cot\beta)$ apart.

If instead the cars are on the same side, you subtract: $|x_A - x_B| = H|\cot\alpha - \cot\beta|$ .

Cloud and its reflection

A classic problem: a cloud is at height $h$ above a lake. An observer is at height $a$ above the lake. The angle of elevation of the cloud is $\alpha$ and the angle of depression of its reflection in the lake is $\beta$ ( $\beta > \alpha$ ).

The reflection appears at depth $h$ below the lake surface (mirror image). If the horizontal distance from observer to cloud is $d$ :

$\tan\alpha = (h-a)/d, \qquad \tan\beta = (h+a)/d.$

Eliminating $d$ : $(h-a)\cot\alpha = (h+a)\cot\beta$ , giving

$h = a \cdot \frac{\tan\beta + \tan\alpha}{\tan\beta - \tan\alpha}.$

This formula is asked verbatim in many board papers.

Worked examples

Example 1. From the top of a $7$ -m building, the angle of elevation of the top of a cable tower is $60^\circ$ and the angle of depression of the foot of the tower is $45^\circ$ . Find the height of the tower.

Building height $b = 7$ . Using $d = b/\tan 45^\circ = 7$ , and $h - 7 = d\tan 60^\circ = 7\sqrt 3$ , we get $h = 7 + 7\sqrt 3 = 7(1 + \sqrt 3) \approx 19.12$ m.

Example 2. A statue $1.6$ m tall stands on top of a pedestal. From a point on the ground, the elevation of the top of the statue is $60^\circ$ and the elevation of the top of the pedestal is $45^\circ$ . Find the height of the pedestal.

Let the pedestal height be $p$ and the horizontal distance be $x$ .

$\tan 45^\circ = p/x \Rightarrow p = x$ .
$\tan 60^\circ = (p + 1.6)/x = \sqrt 3 \Rightarrow p + 1.6 = x\sqrt 3 = p\sqrt 3$ .
So $p(\sqrt 3 - 1) = 1.6 \Rightarrow p = 1.6/(\sqrt 3 - 1) = 1.6(\sqrt 3 + 1)/2 = 0.8(\sqrt 3 + 1) \approx 2.19$ m.

Example 3. A boy $1.5$ m tall is $30$ m from a building. He sees the top of the building at $60^\circ$ above the horizontal from his eye. Find the building's height.

$\tan 60^\circ = (H - 1.5)/30 \Rightarrow H - 1.5 = 30\sqrt 3 \Rightarrow H = 1.5 + 30\sqrt 3 \approx 53.46$ m.

Example 4. From two points $A$ and $B$ on the ground, on opposite sides of a $50$ -m tower, the angles of elevation of the top are $30^\circ$ and $60^\circ$ . Find the distance $AB$ .

$x_A = 50/\tan 30^\circ = 50\sqrt 3$ , $x_B = 50/\tan 60^\circ = 50/\sqrt 3 = 50\sqrt 3/3$ .

$AB = x_A + x_B = 50\sqrt 3 + 50\sqrt 3/3 = 200\sqrt 3/3 \approx 115.47$ m.

Example 5. A vertical tower $50$ m tall is on level ground. A wire is stretched from the top of the tower to a point $30$ m from its foot. Find the length of the wire and the angle it makes with the ground.

Length: $\sqrt{50^2 + 30^2} = \sqrt{3400} = 10\sqrt{34} \approx 58.3$ m.

Angle: $\tan\theta = 50/30 = 5/3 \Rightarrow \theta = \arctan(5/3) \approx 59.04^\circ$ .

(When standard angles are not the answer, board questions usually give simple Pythagorean triples or ask for $\tan\theta$ , not the angle itself.)

Try it yourself

From a building $20$ m high, the angles of elevation and depression of the top and bottom of a tower are $60^\circ$ and $30^\circ$ . Find the tower's height.
A statue $4$ m tall is on a pedestal. From a point on the ground, the bottom and top of the statue make elevations $45^\circ$ and $60^\circ$ . Find the pedestal's height.
Two pillars of equal height stand on either side of a $100$ -m wide road. From a point on the road, the elevations of the tops are $30^\circ$ and $60^\circ$ . Find the pillars' height.
From a $60$ -m hill, the depressions of two cars on opposite sides are $30^\circ$ and $45^\circ$ . Find the distance between the cars.
A man at the top of a $30$ -m tower sees a bird flying horizontally at his eye level. The bird's elevation from a point on the ground $20$ m from the tower is $45^\circ$ . Find the bird's height.
A tree breaks and the broken part touches the ground making a $30^\circ$ angle. The foot of the tree is $5$ m from the point where the top touches. Find the original height of the tree.
From a window of a house $4$ m above the ground, the elevation of the top of a tower opposite is $60^\circ$ and the depression of its foot is $30^\circ$ . Find the tower's height.
From the top of a $50$ -m lighthouse, the depressions of two boats on opposite sides are $45^\circ$ and $60^\circ$ . Find the distance between the boats.
A flagstaff $5$ m tall stands on a tower. From a point on the ground, the tower and its top with flagstaff subtend $45^\circ$ and $60^\circ$ . Find the tower's height.
A pole on a building $20$ m tall has its top making an elevation of $45^\circ$ from a point on the ground $20$ m from the foot. Find the pole's height.

Pitfalls / Insight

For "tree breaks" problems, a common framing: a tree of height $H$ breaks at a point, the broken part falls without detaching at the break, its tip touches the ground. If the broken part makes angle $\theta$ with the ground and its tip is $x$ metres from the foot, then the broken part has length $x/\cos\theta$ , the standing part has height $x\tan\theta$ , and $H = x\tan\theta + x/\cos\theta = x(\tan\theta + \sec\theta)$ . Memorise this form , it saves time.

When triangles share a side, draw the shared side first and build outward. If two triangles share an angle but not a side, look for an alternate-angle or vertically opposite-angle argument to relate them.