A boy 1.2m tall, stands 6m away from the foot of a vertical lamp pole 4.2m long. If the lamp is at the tip of the pole, (a) represent this information in a ...
Question 1 Report
A boy 1.2m tall, stands 6m away from the foot of a vertical lamp pole 4.2m long. If the lamp is at the tip of the pole,
(a) represent this information in a diagram ;
(b) calculate the (i) length of the shadow of the boy cast by the lamp ; (ii) angle of elevation of the lamp from the boy, correct to the nearest degree.
Answer Details
(a) Take the pole vertical with the lamp \(L\) at its top \((4.2\,\text{m})\). The boy \(BC\) is \(1.2\,\text{m}\) tall standing \(6\,\text{m}\) from the pole's foot; his shadow \(CS\) stretches away from the pole on the ground. The light ray from \(L\) passes over the boy's head to the shadow tip \(S\).
(b)(i) Let the shadow length be \(s\). By similar triangles (light from \(L\) grazing the boy's head): \[\frac{\text{lamp height}}{\text{pole-to-shadow-tip}} = \frac{\text{boy height}}{\text{boy-to-shadow-tip}} \Rightarrow \frac{4.2}{6 + s} = \frac{1.2}{s}.\] \[4.2 s = 1.2(6 + s) \Rightarrow 4.2s = 7.2 + 1.2s \Rightarrow 3s = 7.2 \Rightarrow s = 2.4.\] Length of the boy's shadow \(= \mathbf{2.4\,\text{m}}.\)
(b)(ii) Angle of elevation of the lamp from the boy (measured from the top of his head): the horizontal distance is \(6\,\text{m}\) and the lamp is \(4.2 - 1.2 = 3.0\,\text{m}\) above him. \[\tan\theta = \frac{3.0}{6} = 0.5 \Rightarrow \theta = 26.57^\circ \approx \mathbf{27^\circ}.\]
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