Asked by Anonymous
If a pilot wants to travel 400 mph at a bearing of 80, what should he set his speed and bearing to counteract a wind blowing at 50 mph coming directly from South?
Answers
Answered by
Reiny
"bearings" to indicate direction is often misunderstood so you should stick with the conventional method used in trigonometry
I will use vectors, so ...
(400cos10° , 400sin10°) = R + (50cos90° , 50sin90°)
where R is the required vector
(393.923, 69.459) = (x,y) + (0, 50)
R = (x,y) = (393.923 , 19.459)
|R| = √(393.923^2 + 19.459^2) = appr 394.4 mph
direction angle θ :
tanθ = 19.459/393.923
θ = appr 2.8°
so his bearing should be 87.2°
check my arithmetic
I will use vectors, so ...
(400cos10° , 400sin10°) = R + (50cos90° , 50sin90°)
where R is the required vector
(393.923, 69.459) = (x,y) + (0, 50)
R = (x,y) = (393.923 , 19.459)
|R| = √(393.923^2 + 19.459^2) = appr 394.4 mph
direction angle θ :
tanθ = 19.459/393.923
θ = appr 2.8°
so his bearing should be 87.2°
check my arithmetic
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