Nazinia is fly fishing in a stream. The tip of her fishing rod is 1.8m above the surface of the water and flies at the end of the string rests on the water 3.6m away and 2.4m from a point directly under the tip of the rod. Assuming that her string(from the tip of her rod to the fly) is taut, how much string does she have out(see given figure)? If she pulls the string at the rate of 5cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Nazinia is fly fishing in a stream. The tip of her fishing rod is 1.8m above the surface of the water and flies at the end of the string rests on the water 3.6m away and 2.4m from a point directly under the tip of the rod. Assuming that her string(from the tip of her rod to the fly) is taut, how much string does she have out(see given figure)? If she pulls the string at the rate of 5cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Hint: Using the given data draw a diagram and use trigonometric ratios and Pythagoras theorem to obtain the solution.

Complete step-by-step answer:

In $\vartriangle ABC$

We apply pythagoras theorem to find out the length of string(AC)

$AC = \sqrt {A{B^2} + B{C^2}} $

$AC = \sqrt {{{(2.4)}^2} + {{(1.8)}^2}} $

$AC = \sqrt {5.76 + 3.24} $

$AC =\sqrt 9 = 3m $

Rate of pulling the string is $5cm/\sec$.

Length of string pulled in 12sec is $12 \times 5 = 60cm = 0.6m$

Now length of remaining string is $3 - 0.6 = 2.4m$

By pulling the string a new triangle is formed $\vartriangle ABD$.

$AD = 2.4m$ and $AB = 1.8m $

Now let's find BD by pythagoras theorem

$ BD = \sqrt {A{D^2} - A{B^2}} $

$ BD = \sqrt {{{(2.4)}^2} - {{(1.8)}^2}} $

$ BD = \sqrt {5.76 - 3.24} = \sqrt {2.52} = 1.58m $

$\therefore$ Horizontal distance of the fly from Nazinia after 12sec =1.58 + 1.2=1.78m.

Note: - In these types of questions always try to do with Pythagoras theorem and trigonometric ratio to find the unknown lengths.