Question

BRAINIEST QUESTION♥️♥️♥️•nazima is fly fishing in a stream.the tip 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 2.4m from a point directly under the tip of the rod assuming that her string is taut,how much string does she have out?if she pulls in the string at rate of 5cm per second, what will be the horizontal distance of the fly from her after 12 seconds?MOST BRAINIEST QUESTION!!!!TRY IF YOU CAN ANS RIGHTLY.NO SPAM ❎❎GOOD LUCK BEST ANSWER WILL BE MARKED AS BRAINIEST 30 POINTS.

Answer 1

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.)

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Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: QUESTION \: \: } \mid}}}}}$

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

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\mathfrak{Let,} \\ \: \: \: \: \: \: \: \: \: \text{ \red{AB} be the height of the rod tip from the} \\ \text{surface of water and \red{BC} be the horizontal } \\ \text{distance between fly to tip of the rod. Then, } \\ \text{\red{AC} will be the length of the string.}$

$\: \: \: \: \: \: \: \underline{ \text{ \: In Figure. 1, \: }} \\ \\ \underline{\bold{ \: \: By \: \: using \: \: Pythagoras \: \: theorem, \: \: in \: \:}} \\ \underline{ \bold{ \: the \: \red{ \triangle \: ABC}, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \bold{AC {}^{2} = AB {}^{2} + BC {}^{2}} \\ \\ \bold{\implies AC {}^{2} = (1.8 \: m) {}^{2} + (2.4 \: m) {}^{2} } \\ \\ \bold{ \implies AC {}^{2} = (3.24 + 5.76) \: m {}^{2} } \\ \\ \bold{\implies AC {}^{2} = 9 \: m {}^{2} } \\ \\ \bold{\implies AC = \sqrt{ 9 \: m {}^{2} } } \\ \\ \: \: \: \: \: \: \bold{ \therefore \: \red{AC = 3 \: m }} \\ \\ \bold{ \therefore \: The \: \: length \: \: of \: \: the \: \: string, \red{AC \: = 3 \: m}.}$

$\: \: \: \: \: \: \tt{ If \: \: Nazima \: \: pulls \: \: in \: \: the \: \: string \: \: at \: \: the \: \: } \\ \tt{rate \: \: of \: \: 5 cm/s, \: then \: \: the \: \: distance \: \: } \\ \tt{ travelled \: \: by \: \: fly \: \: in \: \: 12 \: \: seconds \: \: will \: \: be } \\ \\ \tt{= \red{( 5 \times 12) \: cm }= \red{60 \: cm} = \red{0.6 \: m}}$

$\mathfrak{Let,} \\ \: \: \: \: \: \: \: \: \text{ \red{D} be the position of fly after \red{12 seconds}. } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text{[Figure. 2]} \\ \text{ Hence, } \\ \: \: \: \text{ \red{AD} will be the length of string which is out} \\ \text{ after \red{12 seconds}.} \\ \\ \tt{ \therefore \: Length \: of \: \: the \: \: string \: \: pulls \: \: by \: \: Nazima} \\ \\ \: \: \: \: \: \: \: \: \: \tt{ = \red{AD} = \red{3 \: m - 0.6 \: m} = \red{2.4 \: m}}$

$\mathfrak{Now,} \\ \: \: \: \: \: \: \: \: \: \: \underline{ \text{ \: In \: Figure. 2, \: }} \\ \\ \underline{ \bold{ \: By \: \: using \: \: Pythagoras \: \: theorem \: \: in \: \: }} \\ \underline{ \bold{ \: the \: \triangle ADB, \: }} \\ \\ \: \: \: \: \: \: \: \: \: \: \bold{AD {}^{2} = AB {}^{2} + BD {}^{2} } \\ \\ \bold{ \implies (2.4 \: m) {}^{2} =(1.8 \: m) {}^{2} + BD {}^{2} } \\ \\ \bold{\implies 5.76 \: m {}^{2} = 3.24 \: m {}^{2} + BD {}^{2} } \\ \\ \bold{\implies BD {}^{2} = (5.76 - 3.24) \: m {}^{2} } \\ \\ \bold{\implies BD {}^{2} = 2.52 \: m {}^{2}} \\ \\ \bold{ \implies \: BD = \sqrt{2.52 \: m {}^{2} }} \\ \\ \: \: \: \: \: \bold{\therefore \: \red{BD = 1.59 \: m }\: \: \: (Approx.)}$

$\sf{Hence, \: the \: \: horizontal \: \: distance \: \: of \: \: the \: \: fly \: \: } \\ \sf{ from \: \: Nazima \: \: after \: \: 12 \: seconds \:= (1.59 + 1.2 ) \: m } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{= \pink{2.79 \: m} \: (Approx.)}$