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Chapter 9 · Question 2

A tower stands vertically on the ground. From a point on the ground, which is 15 m15\text{ m} away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60∘60^\circ. Find the height of the tower.

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Q2

A tower stands vertically on the ground. From a point on the ground, which is 15 m15\text{ m} away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60∘60^\circ. Find the height of the tower.

Answer Revealed
Direct Answer:
Let ABAB be the tower and CC be the observation point. In right △ABC\triangle ABC, tan⁡60∘=ABBC\tan 60^\circ = \frac{AB}{BC}. With tan⁡60∘=3\tan 60^\circ = \sqrt{3} and BC=15 mBC = 15\text{ m}, we get 3=AB15\sqrt{3} = \frac{AB}{15}, so AB=153 mAB = 15\sqrt{3}\text{ m}.

Simple Explanation

The tower height is 15315\sqrt{3} metres. Since the tangent of 60∘60^\circ equals height divided by the 15 m15\text{ m} ground distance, we multiply to get approximately 15×1.732≈25.98 m15 \times 1.732 \approx 25.98\text{ m}.

Exam-Ready Structure

Let ABAB represent the height of the tower and CC be the observation point on the ground. In right-angled triangle ABCABC (right-angled at BB): tan⁡60∘=ABBC\tan 60^\circ = \frac{AB}{BC}. Given: BC=15 mBC = 15\text{ m} (distance from foot of tower). tan⁡60∘=3\tan 60^\circ = \sqrt{3}. Substituting: 3=AB15\sqrt{3} = \frac{AB}{15}. Therefore, AB=153 mAB = 15\sqrt{3}\text{ m}. Hence, the height of the tower is 153 m15\sqrt{3}\text{ m} (approximately 15×1.732=25.98 m15 \times 1.732 = 25.98\text{ m}). Note: We chose tan⁡\tan because it relates the known adjacent side (BCBC) and the unknown opposite side (ABAB).

Key Points

  • Draw right triangle with tower as perpendicular and ground distance as base
  • Use tan⁡θ=oppositeadjacent\tan \theta = \frac{\text{opposite}}{\text{adjacent}}
  • tan⁡60∘=3\tan 60^\circ = \sqrt{3}, so h=153 mh = 15\sqrt{3}\text{ m}
  • Choose the ratio that connects the known side and the unknown side

Related Questions

Q1

Define the terms (i) angle of elevation and (ii) angle of depression. What is a line of sight?

Q3

An electrician has to repair an electric fault on a pole of height 5 m5\text{ m}. She needs to reach a point 1.3 m1.3\text{ m} below the top of the pole. What should be the length of the ladder that she should use which, when inclined at an angle of 60∘60^\circ to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (Take 3=1.73\sqrt{3} = 1.73)

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