Answer:
From \[\Delta \,ABC,\frac{AB}{BC}=\tan \,\,60{}^\circ \] or \[BC=\frac{AB}{\tan \,\,60{}^\circ }\] \[BC=\frac{150}{\sqrt{3}}m\] From \[\Delta \,ABD,\frac{AB}{BD}=\tan \,\,45{}^\circ \] or \[AB=BD[\because \,tan\,\,45{}^\circ =1]\] \[\Rightarrow \] \[BD=150\,m\] Distance covered in 2 min \[=BD-BC\] \[=150-\frac{150}{\sqrt{3}}=\frac{150\sqrt{3}-150}{\sqrt{3}}\] Distance covered in 1 hour \[=\frac{150\left( \sqrt{3}-1 \right)}{\sqrt{3}\times 2}\times 60\,\,m\] \[speed=\frac{4500(\sqrt{3}-1)}{\sqrt{3}}\] \[=4500-1500\sqrt{3}\] \[=4500-2598=1902\text{ }m/hr.\] Hence, the speed of boat is \[1902\text{ }m/hr\].
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