Chapter 9 · Question 4

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

Answer Revealed

Direct Answer:

Let the chimney be

AB

and observer

CD

of height

1.5\text{ m}

. In right

\triangle ADE

\tan 45^\circ = \frac{AE}{DE} = \frac{AE}{28.5}

. Since

\tan 45^\circ = 1

AE = 28.5\text{ m}

. Height of chimney

AB = AE + BE = 28.5 + 1.5 = 30\text{ m}

Simple Explanation

Because the angle of elevation is

45^\circ

, the extra height above eye level equals the ground distance (

28.5\text{ m}

). Add the observer's height (

1.5\text{ m}

) to get the total chimney height of

30\text{ m}

Exam-Ready Structure

Let

AB

represent the chimney and

CD

the observer (

CD = 1.5\text{ m}

). Draw a horizontal line from the observer's eye

D

to the chimney at

E

, forming right

\triangle ADE

. In

\triangle ADE

\angle ADE = 45^\circ

(angle of elevation),

DE = CB = 28.5\text{ m}

. Using

\tan 45^\circ

\tan 45^\circ = \frac{AE}{DE}

. Since

\tan 45^\circ = 1

, we get

\frac{AE}{28.5} = 1

, so

AE = 28.5\text{ m}

. Now

AB = AE + EB

, and

EB = CD = 1.5\text{ m}

(height of observer). Therefore,

AB = 28.5 + 1.5 = 30\text{ m}

. The height of the chimney is

30\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: $AE = DE \times \tan 45^\circ = 28.5\text{ m}$
Add observer's height: $28.5 + 1.5 = 30\text{ m}$
For $45^\circ$ , the height above eye level equals the ground distance ( $\tan 45^\circ = 1$ )

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