1. Wendy leaves airport A, flying on a bearing of 325 degrees for 800km.
She then turns on a course of 235 degrees and flies for 950km. Find Wendy's bearing from A. frown
2. Wendy decides to turn and fly straight back to A at a speed of 450 km/h. How long will it take her?
There are two coordinate systems used to solve this problem. The first system, used in aeronautics,
is a system in which angles are measured clockwise from a line that runs from south to north.
The second, used to plot graphs, measures angles counter-clockwise from the +X axis, which
runs from left to right. It is necessary to convert the angles stated in the problem to equivalent
angles in the rectangular coordinate system.
Bearing 325 degrees is 35 degrees to the left of north. North converts to the positive Y axis, so the
angle we want is the complementary angle measured up from the negative X axis.
90 - 35 = 55
If we could show a graph, we would see a line extending up to the left, from the origin at (0,0) to
a point 800Km (scaled for the graph) from the origin. What we want to know is the X and Y coordinates
of that point. By dropping a perpendicular line down to the X axis, we form a right-triangle, with
the flight path forming the hypotenuse. We'll call the 55 degree angle at the origin angle A, and the
angle at the far end of the flight path angle B. Keep in mind that angle B is the complement of angle A,
so it is 35 degrees. Because we will be working with more than one triangle, let's make sure we can
differentiate between the x and y sides of the triangles by including numbers with the tags.
Side y1 is opposite the 55 degree angle, so...
y1 / 800 = sin 55
y1 = (sin 55) * 800
y1 = 0.8191 * 800 = 655.32 We'll round that down, to 655 Km
Side x1 is opposite the 35 degree angle at the top, so...
x1 / 800 = sin 35
x1 = (sin 35) * 800
x1 = 0.5736 * 800 = 458.86 We'll round that up, to 459 Km
At this point, the aircraft turns to a heading of 235 degrees. Due west is 270 degrees, so the new
heading is 35 degrees south of a line running right to left, which is the negative X axis. Temporarily,
we move the origin of the rectangular coordinate system to the point where the turn was made, and
proceed as before. We draw a line 950Km down to the left, at a 35 degree angle. From the far endpoint
of that line, we drop a perpendicular line to the -X axis ("drop" is the term even though we can see
that the axis is above the flight path).
Side y2 is opposite this triangle's 35 degree angle at the adjusted origin, so...
y2 / 950 = sin 35
y2 = (sin 35) * 950
y2 = 0.5736 * 950 = 544.89 We'll round that up, to 545 Km
Side x2 is opposite this triangle's 55 degree angle, so...
x2 / 950 = sin 55
x2 = (sin 55) * 950
x2 = 0.8191 * 950 = 778.19 We'll round that down, to 778 Km
The problem asks for the bearing to that second endpoint from the beginning point, which is where
we set the first origin. We now draw our third triangle with a hypotenuse from the origin
to the second endpoint and its own x and y legs.
Because the second flight continued going further out on the negative X axis, we can add
the two x values we calculated above.
x3 = x1 + x2 = 459 Km + 778 Km = 1237 Km
The first leg of the flight was in a northerly direction, but the second leg was in a southerly
direction, meaning that the final endpoint was closer to our -X axis. For that reason, it is
necessary to subtract the second y value from the first y value to obtain the y coordinate
for the triangle we are constructing.
y3 = y1 - y2 = 655 Km - 545 Km = 110 Km
Using x3 and y3, we can determine the angle (let's call it angle D) at the origin by finding the tangent.
tan D = y3 / x3 = 110 / 1237 = 0.0889
Feeding that value into the inverse tangent function, we find the angle that it defines.
tan^-1 0.0889 = 5.08 degrees
The angle we found is based on the rectangular coordinate system. We need to convert that
to the corresponding bearing that was asked for in the problem. We know that the second
endpoint is still to the left of the origin. The -X axis represents due west, or 270 degrees.
We know that the endpoint is above the -X axis, so we must increment the bearing by the size of
the angle we calculated.
Bearing = 270 + 5.08 approximately 275 degrees
The second part of the problem asks how long it will take Wendy (the pilot) to fly
straight back to the origin.
Distance = sqrt (x3^2 + y3^2) = sqrt (778^2 + 1237^2) = sqrt (605284 + 1530169)
= sqrt (2135453) = 1461.319 We'll round that one, too 1461 Km
Wendy will fly 1461 Km at 450 Km/hr. How long will that take?
t = d / s = 1461Km / (450Km/hr) = 3.25 hours << 3hrs 15 mins
