Nazima is fly fishing in a stream. The tip of her fishing rod is $1.8 \, m$ above the surface of the water and the fly at the end of the string rests on the water $2.4 \, m$ away 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? If she pulls in the string at the rate of $5 \, cm$ per second,what will be the horizontal distance of the fly from her after $12 \, seconds$?

(N/A) Let $A$ be the tip of the fishing rod and $B$ be the point on the water surface directly below $A$. Let $C$ be the initial position of the fly. Given $AB = 1.8 \, m$ and $BC = 2.4 \, m$.
In the right-angled triangle $\triangle ABC$,by Pythagoras theorem:
$AC^2 = AB^2 + BC^2$
$AC^2 = (1.8)^2 + (2.4)^2 = 3.24 + 5.76 = 9.00$
$AC = \sqrt{9} = 3 \, m$.
So,the length of the string out is $3 \, m$.
She pulls the string at a rate of $5 \, cm/s = 0.05 \, m/s$.
In $12 \, seconds$,the length of the string pulled in is $12 \times 0.05 = 0.6 \, m$.
Let the new position of the fly be $D$ after $12 \, seconds$. The new length of the string is $AD = AC - 0.6 = 3 - 0.6 = 2.4 \, m$.
In the right-angled triangle $\triangle ADB$:
$AD^2 = AB^2 + BD^2$
$(2.4)^2 = (1.8)^2 + BD^2$
$5.76 = 3.24 + BD^2$
$BD^2 = 5.76 - 3.24 = 2.52$
$BD = \sqrt{2.52} \approx 1.587 \, m$.
The total horizontal distance of the fly from Nazima is $BD + 1.2 \, m = 1.587 + 1.2 = 2.787 \, m \approx 2.79 \, m$.