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Chapter 9 · Question 4

An observer 1.5 m1.5\text{ m} tall is 28.5 m28.5\text{ m} away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45∘45^\circ. What is the height of the chimney?

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Answer

Direct Answer

Let the chimney be ABAB and observer CDCD of height 1.5 m1.5\text{ m}. In right △ADE\triangle ADE, tan⁡45∘=AEDE=AE28.5\tan 45^\circ = \frac{AE}{DE} = \frac{AE}{28.5}. Since tan⁡45∘=1\tan 45^\circ = 1, AE=28.5 mAE = 28.5\text{ m}. Height of chimney AB=AE+BE=28.5+1.5=30 mAB = AE + BE = 28.5 + 1.5 = 30\text{ m}.

Simple Explanation

Because the angle of elevation is 45∘45^\circ, the extra height above eye level equals the ground distance (28.5 m28.5\text{ m}). Add the observer's height (1.5 m1.5\text{ m}) to get the total chimney height of 30 m30\text{ m}.

Exam-Ready Structure

Let ABAB represent the chimney and CDCD the observer (CD=1.5 mCD = 1.5\text{ m}). Draw a horizontal line from the observer's eye DD to the chimney at EE, forming right △ADE\triangle ADE. In △ADE\triangle ADE, ∠ADE=45∘\angle ADE = 45^\circ (angle of elevation), DE=CB=28.5 mDE = CB = 28.5\text{ m}. Using tan⁡45∘\tan 45^\circ: tan⁡45∘=AEDE\tan 45^\circ = \frac{AE}{DE}. Since tan⁡45∘=1\tan 45^\circ = 1, we get AE28.5=1\frac{AE}{28.5} = 1, so AE=28.5 mAE = 28.5\text{ m}. Now AB=AE+EBAB = AE + EB, and EB=CD=1.5 mEB = CD = 1.5\text{ m} (height of observer). Therefore, AB=28.5+1.5=30 mAB = 28.5 + 1.5 = 30\text{ m}. The height of the chimney is 30 m30\text{ m}. Important: When the observer has height, always add the observer's height to the computed vertical distance above eye level.

Key Points

  • Draw horizontal from observer's eye level to the object
  • Height above eye level: AE=DE×tan⁡45∘=28.5 mAE = DE \times \tan 45^\circ = 28.5\text{ m}
  • Add observer's height: 28.5+1.5=30 m28.5 + 1.5 = 30\text{ m}
  • For 45∘45^\circ, the height above eye level equals the ground distance (tan⁡45∘=1\tan 45^\circ = 1)

Related Questions

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.

Q5

From a point PP on the ground, the angle of elevation of the top of a 10 m10\text{ m} tall building is 30∘30^\circ. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from PP is 45∘45^\circ. Find the length of the flagstaff and the distance of the building from point PP. (Take 3=1.732\sqrt{3} = 1.732)

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