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

"

Given:

Nazima is fly fishing in a stream. The trip of her fishing rod is 1.8m above the surface of the water and the fly at the end of the string rests on the water 3.6m away and 2.4 m from a point directly under the trip of the rod.

To do:

We have to find the length of the string that is out and if she pulls in the string at the rate of 5 cm per second, the horizontal distance of the fly from her after 12 seconds.

Solution:

In $\triangle \mathrm{ABC}$,

$\mathrm{AC}^{2}=\mathrm{BC}^{2}+\mathrm{AB}^{2}$

$=(2.4)^{2}+(1.8)^{2}$

$=5.76+3.24$

$\Rightarrow \mathrm{AC}=\sqrt{9.00}$

$=3 \mathrm{~m}$

Let the length of the string she pulled out in $12 \mathrm{~s}$ be $BD$

This implies,

The new length of the fly $=0.05 \times 12$

$=0.60 \mathrm{~m}$

Remaining length of the fly $(\mathrm{AD})=3-0.60$

$=2.4 \mathrm{~m}$

In $\triangle \mathrm{ABD}$,

$\mathrm{AD}^{2}=\mathrm{AB}^{2}+\mathrm{BD}^{2}$

$\mathrm{BD}^{2}=\mathrm{AD}^{2}-\mathrm{AB}^{2}$

$=(2.4)^{2}-(1.8)^{2}$

$\mathrm{BD}=\sqrt{5.76-3.24}$

$=\sqrt{2.52}$

$=1.59 \mathrm{~m}$

Horizontal length of the fly $=(1.59+1.2) \mathrm{m}$

$=2.79 \mathrm{~m}$

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