A moving boat is observed from the top of a 150 𝑚 high cliff moving away from the cliff. The angle of depression of the boat changes from 60° 𝑡𝑜 45° 𝑖𝑛 2 𝑚𝑖𝑛𝑢𝑡𝑒𝑠. Find the speed of the boat.

Given that,

Height of cliff is 150 𝑚

The angle of depression of the boat changes from 60° 𝑡𝑜 45° 𝑖𝑛 2 𝑚𝑖𝑛𝑢𝑡𝑒𝑠. 

In ∆𝐴𝐵𝐶,

$\begin{array}{r}\tan 60=\frac{AB}{BC} \\ \sqrt{3}=\frac{150}{BC} \\ BC=\frac{150}{\sqrt{3}} \\ BC=\frac{150}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ BC=\frac{150\sqrt{3}}{3} \\ BC=50\sqrt{3}\mathrm{m} \\ \text{ In }\mathrm{△}ABD \\ \tan 45=\frac{AB}{BD} \\ 1=\frac{150}{BC+CD} \\ BC+CD=150\mathrm{m} \\ CD=150\mathrm{m}-50\sqrt{3}\mathrm{m} \\ CD=50(3-\sqrt{3})\mathrm{m} \end{array}$

Time taken to from point 𝐶 to 𝐷 is 2 𝑚𝑖𝑛𝑢𝑡𝑒𝑠

$\begin{array}{r}\Rightarrow \frac{2}{60}=\frac{1}{30}\text{ hours } \\ \text{ Speed }=\frac{\text{ Distance }}{\text{ Time }} \\ \text{ Speed }=\frac{50(3-\sqrt{3})m}{\frac{1}{30}\text{ hours }} \\ \text{ Speed }=1500(3-\sqrt{3})\mathrm{m}/\mathrm{h}\end{array}$

Hence the speed of the boat is $1500 (3 -\sqrt{3})$𝑚/ℎ