Altitude The angles of elevation to an air plane from two points A and B on level ground are 71° and 88°,respectively .The points A and B are 5.5 miles apart and the airplaine is east if both points in the same vertical plane. Find the altitude of the plane.
Find the altitude if the plane
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If the plane and the two points are in the same vertical plane, the solution will be:
Calculate the angle formed by the visual lines at the plane point; It will be 180 - 88 - 71 = 21 degrees
According to the sinus law we may write:
5.5(dist between points)/sin 21 = a(opposite to angle A)/ sin 71 = b/sin 88
solving for a and b we obtain:
a = 14.51 miles
b = 15.33 miles
Then we can form tho rectangles triangles, for one the vertical side, forms 90 degrees with the soil then will be the altitude.
Sinus theorem; 14.1/1 = h/sin 88 solving for h = 14.49 miles
To check this answer we can do it in the other triangle:
15.33/1(sin 90) = h/ sin 71, solving for h gives 14.49. so CHECK!!!!!!!!!!!!
let point C be where the plane is
you have triangle ABC
angle A is 71º
angle B is 88º (outside triangle ABC)
angle B inside the triangle is 92º (180-88)
angle C is 17º (71+92+17=180)
using the law of sines:
now you have another triangle CBD where CD is the altitude of the plane
using the law of sines again:
sin(90)/BC=sin(88)/CD where CD is the altitude
Tan(71°)(x+5.5) = a
Tan(88°)(x) = a
(x)Tan(71°)+(5.5)Tan(71°) = (x)Tan(88°)
(5.5)Tan(71°) = (x)Tan(88°)-(x)Tan(71°)
(5.5)Tan(71°) = x(Tan88°-Tan71°)
x = (5.5)Tan71°
After finding x, plug into: a = (x)Tan88° or a = Tan(71°)(x+5.5) (the former is easier)