9514 1404 393
Answer:
9.5°, yes
Step-by-step explanation:
The relevant trig relation is ...
Tan = Opposite/Adjacent
The distance opposite the angle of elevation is the plane's height, 500 m. The distance adjacent to the angle of elevation is the horizontal distance to the plane, 3 km = 3000 m. Then the angle is found from ...
tan(α) = 500/3000 = 1/6
α = arctan(1/6) ≈ 9.46°
The plane is approaching at an angle of 9.46°. It is safe to land, since that angle is less than 15°.
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<em>Additional comment</em>
The usual descent angle for most commercial air traffic is 3°. Some airport geography demands it be different (steeper). A higher descent angle can put undue stress on the landing gear.
Answer: Hi!
First, UxV = sin(a)*IUI*IVI
where a is the angle between U and V, in this case 45°.
First, the cross product of UxV points:
Here you can use the right hand method,
Put your hand flat, so the point of your fingers point in the same direction that the first vector, in this case U, so your fingers will point to the north.
Now roll your fingers in the direction of the second vector, so here you will roll your fingers in the northeast direction. Now you will see that your thumb is pointing down, so the cross product of UxV points down.
And the magnitude is 6*5*sin(45) = 21.213
Answer:
you need to be more specific like which ones are x or y
Step-by-step explanation:
ask it again or comment the specifics
