Heights and Distances

360. Pole & Tower Heights · Unlocking distances with angles of elevation

The tower height always equals pole height plus the tangent-proportioned upper segment.

BCDE30°60°Pole height = 20 mPole height = 20 mTower height = 80 mTower height = 80 mHorizontal = 34.6 mHorizontal = 34.6 mA
With a pole of known height as reference, angles of elevation to the top and bottom of a tower give two tangent relations. The tower height = pole height + the extra upper segment found from the tangent of the top angle.

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Selina ICSE: Heights and Distances

What this lesson covers

Try to break it

Drag A up or down to change the pole's height. The horizontal sight line from A shifts with it, and the tower's height tracks. The angle of elevation from A to the tower's top and the angle of depression from A to the tower's base are fixed by the problem, so the tower height is always (pole height) + d · tan(elevation), where d is the horizontal gap and (pole height) = d · tan(depression).

The proof, step by step

Prove that the tower height equals the pole height plus the tangent-proportioned upper segment.

  • Identify the two right-angled triangles formed: ΔADE where ∠DAE = 30° and DE = pole height = 20 m.
  • Apply tan 30° in ΔADE: tan 30° = DE/AD ⇒ 1/√3 = 20/AD. Solve for the horizontal distance AD = 20√3 m.
  • Apply tan 60° in ΔADC: tan 60° = CD/AD ⇒ √3 = CD/(20√3). Solve for CD = 60 m. Total tower height CE = CD + DE = 60 + 20 = 80 m.

Worked example

A vertical pole of height 15 m and a vertical tower are on the same level ground. From the top of the pole, the angle of elevation of the top of the tower is 45° and the angle of depression of the foot of the tower is 30°. Find the height of the tower.

Let pole height = 15 m. In ΔADE, tan 30° = 15/AD ⇒ AD = 15√3 m. In ΔADC, tan 45° = CD/AD ⇒ CD = 15√3 m. Total height = 15 + 15√3 ≈ 40.98 m.

  • 40.98 m — correct
  • 35.00 m
  • 45.00 m
  • 30.00 m
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