Step-by-Step Solution
Key Concept: Case study on applications of trigonometry (heights and distances).
(a) Find the initial distance of the boat from the foot of the lighthouse when the angle of depression was $45^\circ$. [1 Mark]
$\tan45^\circ = \dfrac{60}{d_1} \Rightarrow 1 = \dfrac{60}{d_1} \Rightarrow d_1 = 60$ m. [1.0 Mark]
(b) Find the new distance of the boat from the foot of the lighthouse when the angle of depression became $60^\circ$. [1 Mark]
$\tan60^\circ = \dfrac{60}{d_2} \Rightarrow \sqrt3 = \dfrac{60}{d_2} \Rightarrow d_2 = \dfrac{60}{\sqrt3} = 20\sqrt3$ m. [1.0 Mark]
(c) Calculate the distance travelled by the boat towards the lighthouse during this observation. [1 Mark]
$d_1 - d_2 = 60 - 20\sqrt3 = 20(3 - \sqrt3)$ m. [1.0 Mark]
(d) If the boat continues to move towards the lighthouse, will the angle of depression increase or decrease? [1 Mark]
As the boat moves closer to the base, the distance $d$ decreases, so $\tan\theta = h/d$ increases, hence the angle of depression increases. [1.0 Mark]
Correct Answer: $\tan45^\circ = \dfrac{60}{d_1} \Rightarrow 1 = \dfrac{60}{d_1} \Rightarrow d_1 = 60$ m. [1.0 Mark] | $\tan60^\circ = \dfrac{60}{d_2} \Rightarrow \sqrt3 = \dfrac{60}{d_2} \Rightarrow d_2 = \dfrac{60}{\sqrt3} = 20\sqrt3$ m. [1.0 Mark] | $d_1 - d_2 = 60 - 20\sqrt3 = 20(3 - \sqrt3)$ m. [1.0 Mark] | As the boat moves closer to the base, the distance $d$ decreases, so $\tan\theta = h/d$ increases, hence the angle of depression increases. [1.0 Mark]