SOLUTION: I have been working on this math problem and I can't seem to figure it out. I was wondering if someone could help me? Please and Thank You!! I would deeply appreciate it!
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-> SOLUTION: I have been working on this math problem and I can't seem to figure it out. I was wondering if someone could help me? Please and Thank You!! I would deeply appreciate it!
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Question 280562: I have been working on this math problem and I can't seem to figure it out. I was wondering if someone could help me? Please and Thank You!! I would deeply appreciate it!
The Sine and Cosine Ratios
A certain jet is capable of a steady 20 degree climb. How much altitude does the jet gain when it moves 1km throughthe air? Answer to the nearest 50m. Found 3 solutions by richwmiller, arKed, Alan3354:Answer by richwmiller(17219) (Show Source):
You can put this solution on YOUR website! what numbers do you have ?
two numbers!
20 degrees and 1 km
1 km is the hypotenuse and we want to know the vertical leg of the right triangle
20 degrees is the angle opposite the vertical side.
90+20=110
70 is the third angle
You can put this solution on YOUR website! Solving for the altitude:
sine(teta)=opposite(a)/hypotenuse(b)
Substitute the given:(for a distance of 1km)
sin20=a/1000m
By cross multiplication
we get;
a=1000sin20
a=342.02 meters
Substitute the given:(for a distance of 50m)
sin20=a/50m
By cross multiplication
we get;
a=50sin20
a=17.101 meters
You can put this solution on YOUR website! The Sine and Cosine Ratios
A certain jet is capable of a steady 20 degree climb. How much altitude does the jet gain when it moves 1km throughthe air? Answer to the nearest 50m.
-----------------
1 km = 1000 meters = the hypotenuse
Vertical change = 1000*sin(20) = 342 meters
--> 350 meters to the nearest 50
