Math bearings, complex

Question

A plane flies 6km due west of Keira Bay at a constant height of 800 m. Xander sees the plane from

his house 1.6km south of Keira Bay. Find the bearing of the plane from Xander and the angle of the plane floor

Answer 1

Let's put Keira Bay at the origin, with the y axis pointing north and the x axis pointing east.

1 kilometer = 1000 meters. To avoid fractions, let's label the axes in meters.

The plane is to the left of the origin at the point (-6000, 0).

Xander is below the origin at the point (0, -1600).

(These numbers 6000 and 1600 are the lengths of the legs of a right triangle.)

Therefore the straight line from (0, -1600) to (-6000, 0) points to the upper left.

How far to the upper left? In other words, what is the angle between this line and the line that points straight north from Xander? This angle is the angle whose tangent ("opposite over adjacent") is 6000/1600 = 15/4 = 3.75

This angle is about 75 degrees.

In other words, tan(75 degrees) is approximately 3.75

In yet other words, 3.75 is approximately arctan (75 degrees).

So the line from Xander to the plane is about 75 degrees to the left of true north.

To say the same thing in different words, the line from Xander to the plane is about 360-75=285 degrees to the right of true north.

So if you measure your bearings in degrees to the right of true north, the bearing from Xander to the plane is about 280 degrees.

The Pythagorean theorem tells us that the length of the straight line along the ground from the point (0, -1600) to the point (-6000, 0) is

√(1600² + 6000²)

= √(2560000 + 36000000)

= √(38560000)

≅ 6210

So the point (-6000, 0) is approximately 6210 meters away from Xander.

And the plane is in the air 800 meters above the point (-6000, 0).

These numbers 6210 and 800 are the lengths of the legs of another right triangle.

So from Xander's point of view, the angle from the ground up to the plane is the angle whose tangent ("opposite over adjacent") is 800/6210 ≅ 0.12882

This angle is approximately 7.34 degrees

In other words, tan(7.34 degrees) is approximately 0.12882

In yet other words, arctan(0.12882) is approximately 7.34 degrees.

Therefore Xander sees the plane at an angle of approximately 7.34 degrees above the horizon. I hope he's not alarmed.

Answer 2

plane is 6 km west of the bay

Xander sees the plane when 1.6 km south of the bay

bearing = 90 - tan^-1(4/15) = about 90-14.83

= 75.17 W

plane is 800 m = .8 km high, above the ground

angle of the plane with the ground, seen from Xander's position =

tan^-1(.8/1.6)= tan^-1(.5) =

about 26.57 degrees