Question

A jet leaves a runway whose bearing in N 22degrees E from the control tower. After flying 5miles, the jet turns 115degrees and flies on a bearing of S 43degrees E for 4 miles. At that time, what is the bearing of the jet from the control tower?

Answer 1

What is the final answer? I have to round to the nearest degree

Answer 2

The bearing of the jet from the control tower is?

Answer 3

Are you able to figure it out?

Answer 4

Sorry. Not until now

Answer 5

I am sure it requires cosine rule but no result till now

Answer 6

I can see in the example that they do something with cos sin and tan.

Answer 7

Do you have an asnwer to this questio but want to know hoe to do it?

Answer 8

I don't have an answer for it I am still waiting on someone to answer it for me

Answer 9

Ok, well I was workin on the vector on when it takes off, then subtracting it from when it turns and then find the angle then

Answer 10

Well as you know the plane took off with a bearing of 22 degrees. That means 22 degrees from North, or if you start from the poitive x axis, 68 degrees

Anyways, You can wirte a vector as the |magnitude| times <cos angle with the positive x axis, sin angle it makes with the positive x-axis>

So the magnitude is 5mph or 5<cos68, sin68>

SO plane takeoff : <1.87,4.64>

Now when it turned, it turned 115 or 68-115=-47 or 43 degrees below the x-axis which is just, if you added 360 to get a coterniminal (anle wih the same sin and cos
) you get 317 degrees and it was going 4 mph so it would be

4<cos317, sin317>
<2.93, -2.73>

Now we have to subtract the latter vector form the former to get the new vector (the planes new direction

We get <1.87, 4.64>-<2.93, -2.73>= <-1.06, 7.37>

Now we have to find the angle it makes with the positive x-axis ad then put it in perspective of the bearing (the bearing is not the same as the angle the vector makes witht the x-axis the bearing deals with the ale with the y-axis

Anyways to get it we have to do \[\Huge \tan^{-1} \frac{ 7.37 }{ -1.06 }\]=\[\Huge -81.82~degrees\]

Answer 11

now this is in the 4th quadran