− 50 67 +1.5−100%
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The rate of the kayak in still water is 3.5 mph
<h3><u>Solution:</u></h3>Given that It takes 3 hours to paddle a kayak 12 miles downstream
Distance covered in downstream = 12 miles
Time taken to cover downstream = 3 hours
Also given that it takes 4 hours for the return trip upstream
Distance covered in upstream = 12 miles
Time taken to cover upstream = 4 hours
<em><u>Formula to remember:</u></em>
If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then: Speed downstream = (u + v) km/hr and Speed upstream = (u - v) km/hr
Let the speed of Kayak in still water = x mph
And The speed of current = y mph
For downstream:
Speed downstream = x + y
We know that
x + y = 4 ------ eqn 1
<em><u>For upstream:</u></em>
Speed upstream = x - y
x - y = 3 -------- eqn 2
Now let us eqn 1 and eqn 2
Add eqn 1 and eqn 2
x + y + x - y = 4 + 3
2x = 7
x = 3.5
speed of Kayak in still water = x mph = 3.5 mph
Answer:
Step-by-step explanation:
So, on substituting all these values, we get
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<h2>ADDITIONAL INFORMATION :-</h2>Sign of Trigonometric ratios in Quadrants
- sin (90°-θ) = cos θ
- cos (90°-θ) = sin θ
- tan (90°-θ) = cot θ
- csc (90°-θ) = sec θ
- sec (90°-θ) = csc θ
- cot (90°-θ) = tan θ
- sin (90°+θ) = cos θ
- cos (90°+θ) = -sin θ
- tan (90°+θ) = -cot θ
- csc (90°+θ) = sec θ
- sec (90°+θ) = -csc θ
- cot (90°+θ) = -tan θ
- sin (180°-θ) = sin θ
- cos (180°-θ) = -cos θ
- tan (180°-θ) = -tan θ
- csc (180°-θ) = csc θ
- sec (180°-θ) = -sec θ
- cot (180°-θ) = -cot θ
- sin (180°+θ) = -sin θ
- cos (180°+θ) = -cos θ
- tan (180°+θ) = tan θ
- csc (180°+θ) = -csc θ
- sec (180°+θ) = -sec θ
- cot (180°+θ) = cot θ
- sin (270°-θ) = -cos θ
- cos (270°-θ) = -sin θ
- tan (270°-θ) = cot θ
- csc (270°-θ) = -sec θ
- sec (270°-θ) = -csc θ
- cot (270°-θ) = tan θ
- sin (270°+θ) = -cos θ
- cos (270°+θ) = sin θ
- tan (270°+θ) = -cot θ
- csc (270°+θ) = -sec θ
- sec (270°+θ) = cos θ
- cot (270°+θ) = -tan θ