JOMO 6, Short 5

Geometry Level 4

A man was standing on the top of a $300 \text{m}$ tall tower on coast of Arabic sea. He saw a ship in the sea which was coming towards the coast with a uniform velocity. The angle of depression with which he had to see the ship was $30^{\circ}$ . After 40 seconds that angle of depression was seen to be $45^{\circ}$ . Let the speed of the ship be $x$ m/s. The ship needs $y$ seconds more to reach the coast from the current point . Then the value of $x + y$ can be expressed as $\frac{a\sqrt{3} +b}{c}$ where $c$ is a prime number . Find the value of $a+b+c$ .

The answer is 82.

1 solution

Chew-Seong Cheong
Jan 17, 2015

Let the distance from the coast when the man first saw the ship be $d_0$ and that $40$ seconds later be $d_1$ . Then, we have:

$\quad d_0 = \dfrac {300} {\tan {30^\circ}} = \dfrac {300} {\frac {1}{\sqrt{3}}} = 300\sqrt{3}$ m

Similarly, $\quad d_1 = \dfrac {300} {\tan {45^\circ}} = 300$ m.

Therefore, the speed of the ship: $\quad x = \dfrac {d_0-d_1}{40} = \dfrac {300(\sqrt{3}-1)}{40}$ m/s.

The addition time necessary for the ship to reach the coast:

$\quad y = \dfrac {d_1}{x} = \dfrac {300}{\frac {300(\sqrt{3}-1)}{40}} = \dfrac {40} {\sqrt{3}-1} = \dfrac {40(\sqrt{3}+1)} {3-1} = \dfrac {40\sqrt{3}+40}{2}$

$\quad \Rightarrow a = b = 40\space$ and $\space c = 2\quad \Rightarrow a+b+c = \boxed{82}$

JOMO 6, Short 5

The answer is 82.

1 solution

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