Htykuut

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.

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?

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.

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