Question:

1. Determine the vertical distance to be reached.
The pole is 5 m high and the electrician must work 1.3 m below the top, so the required vertical height from the ground is
$$h = 5 - 1.3 = 3.7\ \text{m}.$$
2. Use the given angle (60°) with the horizontal.
The ladder makes an angle \(\theta = 60^{\circ}\) with the ground.
3. Find the length of the ladder (hypotenuse).
Using the sine ratio:
$$\sin 60^{\circ} = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{h}{L}.$$
Since \(\sin 60^{\circ} = \frac{\sqrt{3}}{2}\) and \(\sqrt{3} \approx 1.73\),
$$\sin 60^{\circ} = \frac{1.73}{2} = 0.865.$$
Hence
$$L = \frac{h}{\sin 60^{\circ}} = \frac{3.7}{0.865} \approx 4.28\ \text{m}.$$
(Exact form: \(L = \frac{3.7 \times 2}{\sqrt{3}} = \frac{7.4}{\sqrt{3}} = \frac{7.4\sqrt{3}}{3}\).)
4. Find the horizontal distance from the foot of the ladder to the pole.
Using the cosine ratio or tangent ratio:
- Cosine method:
$$\cos 60^{\circ} = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{d}{L} \;\Rightarrow\; d = L\cos 60^{\circ} = L \times \frac{1}{2} = \frac{L}{2}.$$
Substituting \(L \approx 4.28\) gives
$$d \approx \frac{4.28}{2} = 2.14\ \text{m}.$$
- Tangent method (check):
$$\tan 60^{\circ} = \frac{\text{opposite}}{\text{adjacent}} = \frac{h}{d} \;\Rightarrow\; d = \frac{h}{\tan 60^{\circ}} = \frac{3.7}{1.73} \approx 2.14\ \text{m}.$$
5. State the final answers.
- Length of ladder required: approximately \(4.28\) m (or \(4.3\) m to one decimal place).
- Distance of the foot of the ladder from the pole: approximately \(2.14\) m (or \(2.1\) m to one decimal place).