Two observation points

A problem with a missing distance

Meena wants the height of a water tower. A fence keeps her away from its foot, so she cannot measure the distance to the tower.

She stands at a point $P$ and measures the angle of elevation of the top: $30°$. Then she walks $40$ m straight towards the tower, to a point $Q$, and measures again: $60°$.

One triangle with one angle has two unknowns: the height $h$ and the distance to the foot. One equation cannot find two numbers. A second angle gives a second triangle and a second equation. Now both unknowns can be found.

The four steps

Call the angle at $P$ by the name $\alpha$ and the angle at $Q$ by the name $\beta$.

1. Draw. Draw the tower $TB$ and the two points $P$ and $Q$ on a line through $B$. Mark both angles and the gap $PQ = d$.
2. Name the unknowns. The height is $h = TB$ and the nearer distance is $x = QB$. Then $PB = x + d$.
3. Write one equation for each triangle. In triangle $TQB$, $\tan\beta = \dfrac{h}{x}$. In triangle $TPB$, $\tan\alpha = \dfrac{h}{x + d}$.
4. Solve. Find $x$ from one equation and put it into the other.

Example A: both points on the same side

Use Meena's numbers: $\alpha = 30°$, $\beta = 60°$ and $d = 40$ m.

1. Triangle $TQB$. $\tan 60° = \dfrac{h}{x}$, so $h = x\sqrt{3}$.
2. Triangle $TPB$. $\tan 30° = \dfrac{h}{x + 40}$, so $h = \dfrac{x + 40}{\sqrt{3}}$.
3. Put the two together. $x\sqrt{3} = \dfrac{x + 40}{\sqrt{3}}$, so $3x = x + 40$ and $x = 20$ m.
4. The height. $h = 20\sqrt{3} \approx 34.64$ m.

Check. The farther distance is $PB = 20 + 40 = 60$ m, and $\dfrac{34.64}{60} = 0.577 = \tan 30°$.

The nearer point sees the larger angle. Standing closer to a tower, you look up more steeply.

Example B: the tower is between the two points

Now the two points are on opposite sides of the tower, $90$ m apart. The angle of elevation is $30°$ from $P$ and $60°$ from $Q$.

1. Each distance. $PB = h\cot 30° = h\sqrt{3}$ and $QB = h\cot 60° = \dfrac{h}{\sqrt{3}}$.
2. They add to $90$ m. $h\sqrt{3} + \dfrac{h}{\sqrt{3}} = 90$, so $\dfrac{4h}{\sqrt{3}} = 90$.
3. Solve. $h = \dfrac{90\sqrt{3}}{4} \approx 38.97$ m.
4. The two distances. $PB = h\sqrt{3} = 67.5$ m and $QB = 22.5$ m. They add to $90$ m, as they should.

A shortcut that checks the answer

A distance on the ground is the height times a cotangent: $x = h\cot\beta$. The cotangent is the reciprocal of the tangent. We know $\cot 30° = \sqrt{3}$, $\cot 45° = 1$ and $\cot 60° = \dfrac{1}{\sqrt{3}}$.

• Same side. $PQ = h\cot\alpha - h\cot\beta$, so $h = \dfrac{d}{\cot\alpha - \cot\beta}$.
• Opposite sides. $PQ = h\cot\alpha + h\cot\beta$, so $h = \dfrac{d}{\cot\alpha + \cot\beta}$.

Check example A: $\dfrac{40}{\sqrt{3} - \tfrac{1}{\sqrt{3}}} = \dfrac{40}{1.1547} = 34.64$ m. It agrees.

Example C: angles that add up to a right angle

A pole stands on level ground. Two points on a line to its foot, on the same side, are $4$ m and $9$ m from the foot. The two angles of elevation add up to $90°$. How tall is the pole?

1. The two tangents. The farther point has the smaller angle $\alpha$, so $\tan\alpha = \dfrac{h}{9}$. The nearer point has $\tan\beta = \dfrac{h}{4}$.
2. Use $\alpha + \beta = 90°$. Then $\tan\beta = \cot\alpha = \dfrac{1}{\tan\alpha}$.
3. Equate. $\dfrac{h}{4} = \dfrac{9}{h}$, so $h^2 = 36$.
4. The height. $h = 6$ m. In general, $h = \sqrt{\text{product of the two distances}}$.

Practise with new numbers

Try it in Python

The program finds the height of a tower seen from two points on one side of it. It also checks the answer against the angle at the farther point.

Common mistakes

• Taking $d$ as the distance to the foot. $d$ is the gap between the two points, not the distance from a point to the tower.
• Swapping the angles. The nearer point has the larger angle.
• Adding when the points are on the same side. On the same side the distances subtract. On opposite sides they add.
• Using a decimal for $\sqrt{3}$ too early. Keep the surd until the last step, then round once.

Remember these

• Two angles from two points give two triangles, two equations and two unknowns.
• The height is the same in both triangles. Equate it.
• Same side: $h = \dfrac{d}{\cot\alpha - \cot\beta}$. Opposite sides: $h = \dfrac{d}{\cot\alpha + \cot\beta}$.
• Check by finding the distance to the foot from each point.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Why can one angle not give the height when the distance to the foot is unknown?

2. A tower is seen at $45°$ from a point $15$ m from its foot. How tall is it?

3. Two points are on one side of a tower. Which point sees the tower top at the larger angle?

4. Two points are $20$ m and $30$ m from the foot of a tower. How far apart are they when they are on the same side, and when they are on opposite sides?

5. Meena walks away from the tower instead of towards it. Does the angle of elevation grow or shrink?