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:
correct answer is 30 degrees
Hi there!
We can use right-triangle trigonometry to solve.
We are given the HYPOTENUSE and ADJACENT sides, so we must use cosine in this instance.
cosθ = Adjacent/Hypotenuse
We can plug in what is given:
cos(28) = A/17
Solve for 'A':
17cos(28) = <u>15.01 ft</u>
Step-by-step explanation:
I think this might help
