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A man on the top of the tower, standing in the seashore finds that a boat coming towards him makes 10 minutes to change the angle of depression from $30$ to $60$. How soon will the boat reach the seashore. enter image description here

My Attempt, We know, Speed $=\frac {dist. }{time }$ $=\frac {CD}{10}$.

Also, $Tan 60=\frac {AB}{BC}$ $\sqrt {3}=\frac {AB}{BC}$ $AB=\sqrt {3} BC$.

Now, what do I have to do further?

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    You need to assume the boat is travelling at a constant velocity. The problem would change if the boat was moving due to a rope which the man in the tower was pulling at a constant rate – 2017-01-25

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If the side length of $CB $ is $x $, then in $\triangle ABC $, we have length of $AB =\sqrt {3}x $. Now in $\triangle ABD $, we have $BD =\sqrt {3}x \cot 30 =3x $. Thus $CD=2x $.

If the boat covers a distance of $2x $ in ten minutes, it can cover a distance of $x $ in what time?? Hope it helps. enter image description here

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    what is $\triangle BD$?? – 2017-01-25
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Let's consider $AB$ to be $h$. Then, $$ BD=\frac{h}{\tan30^\circ}=h\sqrt{3} \\ BC=\frac{h}{\tan60^\circ}=\frac{h}{\sqrt{3}}\\ CD=BD-BC=\frac{2h}{\sqrt{3}} $$ If the speed of the boat is $x$ metres per second, $$ CD=\frac{2h}{\sqrt{3}}=600x \implies \frac{h}{x\sqrt{3}}=300 $$ Time in which the boat travels distance $BC$ is : $$ \frac{BC}{x}=\frac{h}{x\sqrt{3}}=300 $$ Hence, the boat takes $300$ seconds or $5$ minutes to reach the shore. In total, it takes $15$ minutes to cover distance $BD$