Example 5 A vertical pole \( A B \) is erected on a level ground. A man 1.7 m tall stands at \( C, 24 \mathrm{~m} \) away from the foot, \( B \), of the pole. The angle of elevation of the top \( A \) of the pole from the man is \( 54^{\circ} \). Calculate correct to one decimal place, the height of the pole.

To find the height of the pole \( AB \), we can use trigonometry. Let's denote the height of the pole as \( h \). The height from the man's head to the top of the pole is \( h - 1.7 \) m. Using the angle of elevation and the distance from the man to the base of the pole, we can set up the equation: \[ \tan(54^\circ) = \frac{h - 1.7}{24} \] Now, rearranging this to find \( h \): \[ h - 1.7 = 24 \tan(54^\circ) \] \[ h = 24 \tan(54^\circ) + 1.7 \] Calculating \( \tan(54^\circ) \): \[ \tan(54^\circ) \approx 1.3764 \] Now substituting this value into the equation: \[ h = 24 \times 1.3764 + 1.7 \approx 32.4 + 1.7 \approx 34.1 \] Thus, the height of the pole \( AB \) is approximately \( 34.1 \) meters.