Why height and distance problems matter
Ever wondered how engineers figure out the height of a skyscraper without a tape measure? They use the same tricks you’ll master for your ICSE maths exam.
💡 In Simple Words: Height‑and‑distance questions turn real‑world pictures into right‑angled triangles. By using sine, cosine or tangent you can find a hidden length – like the height of a tree – just from an angle and a short measuring stick.
What you need to know first
These problems are all about right‑angled triangles. A right‑angled triangle has one corner that measures 90°, just like the corner of a sheet of paper.
Key trig ratios
Trig ratios link an angle to the sides of a right‑angled triangle. Remember them like this:
| Ratio | Formula | When to use it |
|---|---|---|
| sin (sine) | Opposite ÷ Hypotenuse | You know the side opposite the angle and the longest side. |
| cos (cosine) | Adjacent ÷ Hypotenuse | You know the side next to the angle and the longest side. |
| tan (tangent) | Opposite ÷ Adjacent | You know the two sides that form the right angle. |
Step‑by‑step method
Whenever you see a height‑or‑distance story, follow these four moves:
- Draw the picture – sketch the triangle, label the known angles and sides.
- Choose the right ratio – decide whether sin, cos or tan connects the angle you have with the side you need.
- Write the equation – plug the numbers into the chosen formula.
- Solve for the unknown – rearrange the equation, do the arithmetic, and you’ve got the answer.
Worked example 1: Finding a tree’s height
A boy stands 15 m from a tree and measures the angle of elevation to the top as 30°. How tall is the tree?
Step 1 – Draw: Right‑angled triangle, base = 15 m, angle at the boy = 30°, height = ?
Step 2 – Choose ratio: Height is opposite the angle, base is adjacent, so use tan (opposite/adjacent).
Step 3 – Write equation: tan 30° = height ÷ 15.
Step 4 – Solve: tan 30° ≈ 0.577. So, height = 0.577 × 15 ≈ 8.66 m.
Answer: The tree is about 8.7 m tall.
Worked example 2: Distance across a river
From point A on one bank you spot a lighthouse on the opposite bank. The angle of elevation is 45° and the height of the lighthouse is 20 m. What is the width of the river?
Step 1 – Sketch: Right‑angled triangle, height = 20 m (opposite), width = ? (adjacent), angle = 45°.
Step 2 – Pick ratio: Adjacent side is needed, so use cot (adjacent/opposite) or rewrite tan: tan 45° = opposite ÷ adjacent.
Step 3 – Equation: tan 45° = 20 ÷ width.
Step 4 – Solve: tan 45° = 1, so 1 = 20 ÷ width ⇒ width = 20 m.
Answer: The river is 20 m wide.
Quick tips you can’t forget
- Always label the diagram – a missing label is a missing answer.
- Check whether the side you need is opposite, adjacent, or the hypotenuse.
- If the angle is given as “angle of elevation” or “depression”, draw the line of sight as the hypotenuse.
- Use a calculator for sin, cos, tan values, but keep a few common angles (30°, 45°, 60°) memorised.
- After you get a number, ask yourself: does it look realistic? If a tree comes out 200 m tall, you probably made a mistake.
📝 Likely Exam Questions
- Question: A tower casts a shadow 12 m long when the sun’s elevation angle is 40°. Find the height of the tower.
Answer: Use tan 40° = height ÷ 12 ⇒ height = 12 × tan 40° ≈ 12 × 0.8391 ≈ 10.07 m. - Question: From a point 25 m away from a building, the angle of elevation to the top is 35°. What is the building’s height?
Answer: tan 35° = height ÷ 25 ⇒ height = 25 × tan 35° ≈ 25 × 0.7002 ≈ 17.5 m. - Question: Two observers are 100 m apart on level ground. One sees the top of a hill at 20° above the horizontal, the other at 30°. Find the height of the hill.
Answer: Let the nearer observer be at distance x from the hill base. Then tan 20° = h ÷ x and tan 30° = h ÷ (100‑x). Solve: h = x tan 20° = (100‑x) tan 30°. ⇒ x tan 20° = (100‑x) tan 30°. Solve for x ≈ 61.2 m, then h = 61.2 × tan 20° ≈ 61.2 × 0.3640 ≈ 22.3 m. - Question: A kite is flying at an angle of 55° with the ground. If the string length is 40 m, how high is the kite?
Answer: Height = 40 × sin 55° ≈ 40 × 0.8192 ≈ 32.8 m.