ICSE Class 10 Maths: Trigonometry Height & Distance Word Problems

Why bother with heights and distances?

Ever wondered how engineers figure out the height of a tower without climbing it? Or how a surveyor measures the gap between two hills from the ground? Trigonometry lets you solve those puzzles on paper.

💡 In Simple Words: Height‑and‑distance problems use the angles you see and the basic trig ratios (sine, cosine, tangent) to find hidden lengths. Think of it like using a ruler and a protractor together, but the ruler is invisible.

What are Height & Distance Problems?

These are word problems that ask you to find an unknown height (like a tree) or an unknown distance (like how far a boat is from a lighthouse). The key is that the situation can be drawn as a right‑angled triangle – a triangle with one 90° corner.

Key Trigonometric Ratios to Remember

When you see a right‑angled triangle, three ratios link an angle to the sides:

• sin (θ) = opposite / hypotenuse – opposite means the side across from the angle, hypotenuse is the longest side.
• cos (θ) = adjacent / hypotenuse – adjacent is the side next to the angle, not the hypotenuse.
• tan (θ) = opposite / adjacent – just the two legs, no hypotenuse.

Remember them like a recipe: pick the ratio that contains the side you know and the side you want.

Step‑by‑Step Method

Here’s a quick checklist you can follow for any height‑or‑distance question:

1. Read the problem. Spot the angle you are given and what you need to find.
2. Draw a clean diagram. Mark the right angle, label the known side, the unknown side, and the angle.
3. Choose the right trig ratio (sin, cos or tan) that connects the known side with the unknown side.
4. Write the equation, plug in the numbers, and solve for the unknown.
5. Check the answer – does it make sense? Is it positive and realistic?

Worked Example 1 – Finding the Height of a Tree

Problem: From a point 30 m away from the base of a tree, the angle of elevation to the top is 35°. What is the height of the tree?

Solution:

1. Angle given = 35°. The side we know is the distance from the point to the tree (30 m). The side we want is the height (opposite side).
2. Since we have opposite and adjacent, use tan: tan θ = opposite / adjacent.
3. Write the equation: tan 35° = height / 30.
4. Look up tan 35° (≈ 0.700). So 0.700 = height / 30.
5. Multiply both sides by 30: height = 0.700 × 30 ≈ 21 m.

So the tree is about 21 metres tall. Does that feel right? A 30‑metre walk and a modest angle usually gives a height in the low‑20s.

Worked Example 2 – Distance Between Two Buildings

Problem: From the top of Building A, you look down to the base of Building B and measure an angle of depression of 12°. The height of Building A is 45 m. How far apart are the two buildings?

Solution:

1. An angle of depression is measured downwards from a horizontal line. It equals the angle of elevation from Building B’s base to the top of Building A (alternate interior angles). So the angle we use is 12°.
2. The vertical side (opposite) is the height difference. Since we’re looking from the top of A to the base of B, the opposite side is 45 m.
3. We need the horizontal distance (adjacent). Use tan: tan 12° = opposite / adjacent.
4. tan 12° ≈ 0.213. So 0.213 = 45 / distance.
5. Rearrange: distance = 45 / 0.213 ≈ 211 m.

Hence the buildings are roughly 211 metres apart.

Quick Summary Table

Common Mistakes to Avoid

• Mixing up opposite and adjacent – always label them on your diagram.
• Forgetting to convert the angle to degrees (or radians) as your calculator expects.
• Leaving the answer in a weird unit – keep metres or feet consistent.

📝 Likely Exam Questions

1. Question: A kite is flying 40 m above the ground. From a point 25 m away from the person holding the kite string, the angle of elevation is 58°. Find the height of the kite above the ground.
   Answer: Use tan 58° = height / 25 → tan 58° ≈ 1.600 → height = 1.600 × 25 = 40 m (matches given, confirming consistency).
2. Question: The angle of elevation of a hill’s peak from a point 150 m away is 22°. What is the height of the hill?
   Answer: tan 22° ≈ 0.404 → height = 0.404 × 150 ≈ 60.6 m.
3. Question: From the top of a 30 m high tower, the angle of depression to a boat is 15°. How far is the boat from the base of the tower?
   Answer: tan 15° ≈ 0.268 → distance = 30 / 0.268 ≈ 112 m.
4. Question: Two poles stand on level ground. From the base of the first pole, the angle of elevation to the top of the second pole is 30°. If the first pole is 12 m tall, find the height of the second pole.
   Answer: tan 30° = (height₂‑12) / horizontal distance. Need extra info – typical exam gives distance, e.g., 20 m. Assuming 20 m: tan 30° ≈ 0.577 → 0.577 = (h₂‑12)/20 → h₂‑12 = 11.54 → h₂ ≈ 23.5 m.
5. Question: A surveyor stands 80 m from a cliff base and measures an angle of elevation of 40°. How tall is the cliff?
   Answer: tan 40° ≈ 0.839 → height = 0.839 × 80 ≈ 67.1 m.