ICSE Class 10 Maths: Heights & Distances Trig Problems

Why heights and distances pop up in everyday life

Ever wondered how a surveyor figures out the height of a tree without climbing it? Or how a TV antenna installer knows the exact length of cable to reach the rooftop? Those are classic heights and distances word problems – and they’re all about trigonometry.

💡 In Simple Words: When you know an angle and one side of a right‑angled triangle, you can use trigonometric ratios (sine, cosine, tangent) to find the missing side. That’s the heart of every height‑and‑distance question.

Understanding the basic idea

In these problems we always deal with a right‑angled triangle. One of the legs is the “height” (vertical side) you’re after, the other leg is the “distance” (horizontal side) from your eye or a point on the ground, and the hypotenuse is the line of sight or a rope.

Key trigonometric ratios you’ll use

Remember: opposite means the side across from the angle you’re looking at, adjacent is the side that touches the angle (but isn’t the hypotenuse), and the hypotenuse is the longest side opposite the right angle.

Step‑by‑step method to solve a word problem

Worked example 1

Problem: A student stands 30 m from a flagpole. The angle of elevation from the ground to the top of the pole is 45°. Find the height of the pole.

Solution:

• Draw the triangle: the distance (30 m) is the adjacent side, the height is opposite, and the angle at the student’s eye is 45°.
• Because the angle is 45°, tan 45° = 1. So, tan θ = opposite ÷ adjacent ⇒ 1 = height ÷ 30.
• Multiply both sides by 30: height = 30 m.
• Check: tan 45° = 30/30 = 1 – correct!

Worked example 2

Problem: From a point on the ground, the angle of depression to the base of a lighthouse is 30°. The angle of elevation to the top is 60°. The distance from the point to the lighthouse’s base is unknown. Find the height of the lighthouse.

Solution:

• Draw two right triangles sharing the same horizontal distance (let’s call it d). The lower triangle uses the angle of depression 30°; the upper uses the angle of elevation 60°.
• For the lower triangle, tan 30° = opposite ÷ adjacent = height of eye level ÷ d. We don’t need the eye‑level height; it cancels out later.
• For the upper triangle, tan 60° = height of lighthouse ÷ d. Since tan 60° = √3, we have height = √3 · d.
• From the lower triangle, tan 30° = 1/√3 = eye‑level ÷ d ⇒ eye‑level = d/√3.
• The total height = eye‑level + height above eye = d/√3 + √3 · d = (1/√3 + √3) · d = (4/√3) · d.
• But we still have two unknowns. The trick is to realise the distance d cancels when we use the ratio of the two tangents: (height) / (eye‑level) = tan 60° / tan 30° = √3 / (1/√3) = 3.
• Thus, height = 3 · eye‑level. Substituting eye‑level = d/√3 gives height = 3·(d/√3) = √3 · d, which matches our earlier expression. So any consistent d works; the lighthouse’s height is √3 times the horizontal distance.
• If the problem gave a numeric distance, just plug it in. For example, if d = 50 m, height = √3 · 50 ≈ 86.6 m.

Quick cheat‑sheet

• Identify whether you need sin, cos or tan:
  • Use tan when you know an angle and the two legs (height & distance).
  • Use sin when you know the hypotenuse and the opposite side.
  • Use cos when you know the hypotenuse and the adjacent side.
• Always draw a neat diagram – it saves you from mixing up opposite and adjacent.
• Write the ratio as an equation first, then isolate the unknown.
• Units must match; keep everything in metres or centimetres, not a mix.
• After solving, plug the answer back into the original triangle to see if the angle works.

📝 Likely Exam Questions

1. Question: From a point 40 m away from the base of a tower, the angle of elevation to the top is 30°. Find the height of the tower.
   Answer: Use tan 30° = 1/√3 = height/40 ⇒ height = 40/√3 ≈ 23.1 m.
2. Question: A kite is flying at a height of 60 m. The string makes an angle of 45° with the ground. What is the length of the string?
   Answer: Here, opposite = 60 m, angle = 45°, so sin 45° = 60/length ⇒ length = 60/ (√2/2) = 60·√2 ≈ 84.9 m.
3. Question: Two observation points on level ground are 100 m apart. From the nearer point, the angle of elevation to the top of a building is 20°. From the farther point, it is 10°. Find the height of the building.
   Answer: Let the distance from the nearer point be x. Then height = x·tan 20°. From the farther point, distance = x+100, so height = (x+100)·tan 10°. Equate: x·tan 20° = (x+100)·tan 10°. Solve for x ≈ 277 m. Height = 277·tan 20° ≈ 101 m.
4. Question: A bridge is 80 m long and makes an angle of 30° with the horizontal. Find the vertical height of the bridge above the water.
   Answer: Use sin 30° = opposite/hypotenuse ⇒ 0.5 = height/80 ⇒ height = 40 m.
5. Question: From the top of a building, the angle of depression to a point on the ground 120 m away from the base is 35°. Find the height of the building.
   Answer: Angle of depression equals angle of elevation from ground, so tan 35° = height/120 ⇒ height = 120·tan 35° ≈ 84.0 m.