B
bisector of ZBAC.
10. 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
(see Fig. 6.64)? 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?
2.
Answers
Answer:
2.79 m
Step-by-step explanation:
At first we will have to find the length of AC using pythgoras theorem.
AC² = (2.4)² + (1.8)²
AC² = 5.76 + 3.24 = 9.00AC²
= 5.76 + 3.24 = 9.00
AC = 3 m
length of the string that nazima has out is = 3 m
Given that nazima has pulled the string at the rate of 5 cm/sec in 12 seconds
= (5 x 12) cm = 60 cm = 0.60 m
Remaining string left out = 3 – 0.6 = 2.4 m
Now the length of PB
PB² = PC² – BC²
= (2.4)² – (1.8)²
= 5.76 – 3.24 = 2.52
PB = = 1.59 (approx.)
Hence, after 12 seconds the total horizontal distance of the fly from Nazima is
= 1.59 + 1.2 = 2.79 m (approx.)
Answer:
Ans. I. To find The length of AC.
By Pythagoras theorem,
AC2 = (2.4)2 + (1.8)2
AC2 = 5.76 + 3.24 = 9.00
AC = 3 m
Length of string she has out= 3 m
Length of the string pulled at the rate of 5 cm/sec in 12 seconds
= (5 x 12) cm = 60 cm = 0.60 m
Remaining string left out = 3 – 0.6 = 2.4 m
II. To find: The length of PB
PB2 = PC2 – BC2
= (2.4)2 – (1.8)2
= 5.76 – 3.24 = 2.52
PB = = 1.59 (approx.)
Hence, the horizontal distance of the fly from Nazima after 12 seconds
= 1.59 + 1.2 = 2.79 m (approx.)