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?
"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
To counteract the wind, the pilot needs to adjust the speed and heading of the aircraft. To determine the required adjustments, we need to break down the vectors.
1. Speed Adjustment:
The wind is blowing at 50 mph from the South. Since it is a direct headwind, the effect of the wind on the ground speed is subtractive. Therefore, to travel the desired 400 mph, the pilot needs to increase the aircraft's speed by 50 mph. Thus, the pilot should set the speed to 450 mph.
2. Bearing Adjustment:
The wind is coming from the South, which means it has a bearing of 180 degrees. To counteract the wind's effect, the pilot needs to set a heading that compensates for the drift caused by the wind. The difference between the desired bearing of 80 degrees and the wind's bearing of 180 degrees is 100 degrees. Therefore, the pilot should set the aircraft's heading to 100 degrees.
In summary, the pilot should set the speed to 450 mph and the heading to 100 degrees to counteract the wind and achieve a ground speed of 400 mph at a bearing of 80 degrees.