Chapter 9 · Question 3

An electrician has to repair an electric fault on a pole of height $5\text{ m}$ . She needs to reach a point $1.3\text{ m}$ 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 $60^\circ$ 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 $\sqrt{3} = 1.73$ )

Answer Revealed

Direct Answer:

The point on the pole is at height

5 - 1.3 = 3.7\text{ m}

. Using

\sin 60^\circ = \frac{3.7}{BC}

(ladder length

BC

BC = \frac{3.7 \times 2}{\sqrt{3}} = \frac{7.4}{1.73} \approx 4.28\text{ m}

. Using

\cot 60^\circ = \frac{DC}{3.7}

(distance

DC

DC = \frac{3.7}{\sqrt{3}} = \frac{3.7}{1.73} \approx 2.14\text{ m}

The ladder should be about

4.28\text{ m}

long, and its foot should be placed about

2.14\text{ m}

from the base of the pole.

Let the pole be

AD = 5\text{ m}

. The point to reach is

1.3\text{ m}

below the top, so the height of that point is

BD = 5 - 1.3 = 3.7\text{ m}

. Let the ladder be

BC

and the distance from the pole foot to the ladder foot be

DC

. Finding ladder length ( $BC$ ): In right

\triangle BDC

\sin 60^\circ = \frac{BD}{BC}

i.e.,

\frac{\sqrt{3}}{2} = \frac{3.7}{BC}

. So

BC = \frac{3.7 \times 2}{\sqrt{3}} = \frac{7.4}{1.73} \approx 4.28\text{ m}

. Finding distance ( $DC$ ):

\cot 60^\circ = \frac{DC}{BD}

i.e.,

\frac{1}{\sqrt{3}} = \frac{DC}{3.7}

. So

DC = \frac{3.7}{\sqrt{3}} = \frac{3.7}{1.73} \approx 2.14\text{ m}

. Hence, the ladder should be approximately

4.28\text{ m}

long, and the foot should be placed approximately

2.14\text{ m}

away.

First find effective height: $5 - 1.3 = 3.7\text{ m}$
Use $\sin 60^\circ$ to find ladder length (hypotenuse): $\frac{3.7}{\sin 60^\circ}$
Use $\cot 60^\circ$ to find ground distance: $3.7 \times \cot 60^\circ$
Results: ladder $\approx 4.28\text{ m}$ , distance $\approx 2.14\text{ m}$

Related Questions