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 3.6 m away and 2.4 m 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?

Let AB be the height of rod tip from the surface of water and BC is the horizontal distance between fly to tip of the rod.
Then, the lenght of the string is AC.
In ΔABC, by pythagoras theorem
AC² = AB² + BC²
⇒ AB² = (1.8m)² + (2.4m)²
⇒ AB² = (3.24 + 5.76) m²
⇒ AB² = 9.00 m²
⇒ AB = √9 = 3 m
Hence, the lenght of string, which is out, is 3 m.
If she pulls in the string, at the rate of 5 cm/s, then the distance travelled by fly in 12 seconds
= 12 × 5 = 60 cm = 0.6 m
Let, Dbe the position of string that is out after 12 seconds.
The lenght of the string pulls in by Nazima = AD = AC – 12
= (3.00 – 0.6) m
= 2.4 m
In ΔADB,
AB² + BD² = AD²
⇒ (1.8 m)² + BD² = (2.4 m)²
⇒ BD² = (5.76 – 3.24) m² = 2.52m²
⇒ BD = 1.587 m
Horizontal distance travelled by fly
= BD + 1.2 m
= (1.587 + 1.2) m
= 2.787 m
= 2.79 m.