Basic Trig Question

[Masaru]

I have a difficulty answering Question b) of the following:

Aria is doing the running leg of a triathlon course.
From the start point, A, she runs 800 m at a bearing of 060 degrees.
She then runs 600 m at a bearing of 150 degrees.

a) Calculate Aria's bearing from the start point A to the end point of the run.

b) Aria stays at the point she has reached.
Rob runs 700 m at a bearing of 200 degrees from the start point A.
Calculate Aria's bearing from where Rob has reached.

I can work out question a) as follows:
60 + tan-1(600/800) =097 degrees.

But I have difficulty working out question b).
I know that I could work it out if I found the angle size indicated at "x" on the photo attached, but I have no clue how to find it.

Also I must add to say that this is supposed to be a very basic trigonometric question where you can answer with the basic knowledge of Pythagoras theorem, sin/cos/tan and inverse trig, and it should be worked out applying things like sin/cos rules.

I would much appreciate it if someone can help me with this.

Thank you.

P.S. The textbook says that the answer is 066 degrees, but with no working written on it!

 

[mario99]

I got [imath]66.32^{\circ}[/imath]



The red angle that I marked [imath]= 200^{\circ} - 60^{\circ} - \tan^{-1}\frac{600}{800}[/imath]

Let us call it [imath]\theta[/imath].

After you find [imath]\theta[/imath], use the laws of cosine to find the segment [imath]\text{CD}[/imath]

[imath]CD = \sqrt{\sqrt{800^2 + 600^2}^2 + 200^2 - 2\left(\sqrt{800^2 + 600^2}\right)(200)\cos \theta}[/imath]

Finally, use the laws of sine to find [imath]x[/imath]:

[imath]\displaystyle \frac{\sin x}{\sqrt{800^2 + 600^2}} = \frac{\sin \theta}{CD}[/imath]

 

[Dr.Peterson]

Also I must add to say that this is supposed to be a very basic trigonometric question where you can answer with the basic knowledge of Pythagoras theorem, sin/cos/tan and inverse trig, and it should be worked out applying things like sin/cos rules.

An alternative approach, using only basic trig concepts rather than Law of Cosines, etc., is to find the coordinates of each point relative to A. For example, B is (800 cos(30), 800 sin(30)). Find C and D, then consider the right triangle whose hypotenuse is CD.

 

[mario99]

I cannot see segment [imath]\text{AD}[/imath] clearly. If it is [imath]200 \ \text{m}[/imath], the answer is [imath]66.32^{\circ}[/imath]. If it is [imath]700 \ \text{m}[/imath], the answer is [imath]46.41^{\circ}[/imath].

 

[Dr.Peterson]

I cannot see segment [imath]\text{AD}[/imath] clearly. If it is [imath]200 \ \text{m}[/imath], the answer is [imath]66.32^{\circ}[/imath]. If it is [imath]700 \ \text{m}[/imath], the answer is [imath]46.41^{\circ}[/imath].

b) Aria stays at the point she has reached.
Rob runs 700 m at a bearing of 200 degrees from the start point A.
Calculate Aria's bearing from where Rob has reached.

 

[mario99]

I cannot see segment [imath]\text{AD}[/imath] clearly. If it is [imath]200 \ \text{m}[/imath], the answer is [imath]66.32^{\circ}[/imath]. If it is [imath]700 \ \text{m}[/imath], the answer is [imath]46.41^{\circ}[/imath].

b) Aria stays at the point she has reached.
Rob runs 700 m at a bearing of 200 degrees from the start point A.
Calculate Aria's bearing from where Rob has reached.

P.S. The textbook says that the answer is 066 degrees, but with no working written on it!

The OP claimed that the book says the answer is [imath]66^{\circ}[/imath]. This is impossible if Rob runs [imath]700 \ \text{m}[/imath]. Am I missing something?

 

[mario99]

I think that I got what I was missing. The bearing of Aria from where Bob's final stop is [imath]x + 20^{\circ} = 46.41^{\circ} + 20^{\circ} = 66.41^{\circ}.[/imath]

 

[Masaru]

I think that I got what I was missing. The bearing of Aria from where Bob's final stop is [imath]x + 20^{\circ} = 46.41^{\circ} + 20^{\circ} = 66.41^{\circ}.[/imath]

Thank you so much for your help, mario99.
As I said, this question is supposed to be a basic question where you do not have to use sine or cosine rules.
But it seems to be impossible to work it out without using them...

 

[Masaru]

An alternative approach, using only basic trig concepts rather than Law of Cosines, etc., is to find the coordinates of each point relative to A. For example, B is (800 cos(30), 800 sin(30)). Find C and D, then consider the right triangle whose hypotenuse is CD.

Thank you so much for telling me this alternative method to work out this question.
But this method of using coordinates seems to be even more time consuming, and also I must let you know that this question is supposed to be for those people who have not learned even how to specify a coordinate using sine and cosine.
Having said that, it seems to be impossible to work it out without using sine/cosine rules or coordinates.
Thank you.

 

[Dr.Peterson]

Thank you so much for telling me this alternative method to work out this question.
But this method of using coordinates seems to be even more time consuming, and also I must let you know that this question is supposed to be for those people who have not learned even how to specify a coordinate using sine and cosine.
Having said that, it seems to be impossible to work it out without using sine/cosine rules or coordinates.
Thank you.

How can you be sure there is no other way? And have you tried using coordinates? It really isn't so hard.

Maybe they want you do discover that no special knowledge is needed in order to do that. Clearly it is not a trivial problem for which minimal effort is needed, so you should give it a try.

Please tell us exactly what has been taught, so we can guess what method they might expect. If there have been any similar examples, please show one, so we can see what they did.

 

[Masaru]

How can you be sure there is no other way? And have you tried using coordinates? It really isn't so hard.

Maybe they want you do discover that no special knowledge is needed in order to do that. Clearly it is not a trivial problem for which minimal effort is needed, so you should give it a try.

Please tell us exactly what has been taught, so we can guess what method they might expect. If there have been any similar examples, please show one, so we can see what they did.

AC= Square root of (800^2 + 600 ^2) = 1000.
Let point A be the origin.
The coordinates of point C = (1000cos(90-(60 + tan-1(600/800))), 1000sin(90-(60 + tan-1(600/800))))
= (992.820323..., -119.6152423....)
The coordinates of point D = (700cos(220-90), 700sin(220-90))=(-449.9513268..., -536.2311102...)
Let point E be the point of intersection of the lines through C and D such that the angle CED is 90 degrees and the line segment CE is a vertical line parallel to y-axis and the line segment DE is a horizontal line parallel to x-axis.
DE=992.820323+449.9513268 = 1442.77165...
CE=536.2311102-119.6152423 =416.6158679...
The angle CDE = tan-1(416.6158679/1442.77165)=16.10664333
Therefore, the bearing = 90-16.10664333=73.89335667 degrees.

This answer is wrong even though it is a bit close to the right answer of 66 degrees.
I have no idea where I went wrong, and to me, this method is even harder.

Also, what has been taught is as follows:
1) Pythagoras Theorem of a^2 + b^2 = c^2
2) Trigonometric ratios and meaning of sin/cos/tan
3) Finding the value of a trig ratio
4) Finding angle sizes from the value of a trig ratio
5) Finding unknown lengths using the simple trig ratio of sin/cos/tan
6) Angles of elevation and depression

Thank you.

 

[Masaru]

Regarding:
The coordinates of point D = (700cos(220-90), 700sin(220-90))=(-449.9513268..., -536.2311102...)

Sorry, but I made an error when I was typing.
It should be:
The coordinates of point D = (700cos(90-220), 700sin(90-220))=(-449.9513268..., -536.2311102...)

 

[Dr.Peterson]

Regarding:
The coordinates of point D = (700cos(220-90), 700sin(220-90))=(-449.9513268..., -536.2311102...)

Sorry, but I made an error when I was typing.
It should be:
The coordinates of point D = (700cos(90-220), 700sin(90-220))=(-449.9513268..., -536.2311102...)

I don't see any 220 in the problem. Didn't you mean 200?

 

[Dr.Peterson]

AC= Square root of (800^2 + 600 ^2) = 1000.
Let point A be the origin.
The coordinates of point C = (1000cos(90-(60 + tan-1(600/800))), 1000sin(90-(60 + tan-1(600/800))))
= (992.820323..., -119.6152423....)
The coordinates of point D = (700cos(220-90), 700sin(220-90))=(-449.9513268..., -536.2311102...)
Let point E be the point of intersection of the lines through C and D such that the angle CED is 90 degrees and the line segment CE is a vertical line parallel to y-axis and the line segment DE is a horizontal line parallel to x-axis.
DE=992.820323+449.9513268 = 1442.77165...
CE=536.2311102-119.6152423 =416.6158679...
The angle CDE = tan-1(416.6158679/1442.77165)=16.10664333
Therefore, the bearing = 90-16.10664333=73.89335667 degrees.

I would have found coordinates of B and C one at a time; no inverse tangent is needed in that step. But you have C correct. D is wrong due to the error I mentioned.

Then, to find the length and direction of DC, I would use the right triangle I've added here, which is just what you did:

​

This is the same sort of triangle I'd use to find B and C.

I'm not sure any method could take less work than this. This is also essentially what you will do when you learn about vectors.

 

[mario99]

What professor Dave has given in post #3 is good. But for me I prefer to use the same angles that were given in the problem. In this way, the calculations will be faster and you don't need to think much. Also the errors will decrease.

We have learnt that in polar coordinate system, [imath]x = r\cos \theta \ [/imath] and [imath] \ y = r\sin \theta[/imath]. But when the angle starts from the [imath]y\text{-axis}[/imath], reverse them as:

[imath]x = r\sin \theta[/imath]
[imath]y = r\cos \theta[/imath]

Now you can find the coordinates of the points very quickly and without a lot of thinking. For example, the [imath]x\text{-coordinate}[/imath] of the point [imath]\text{C} \approx 992.82[/imath]

Or

[imath]\text{C}_x = 800\sin 60^{\circ} + 600\sin 150^{\circ} \approx 992.82[/imath]

Try it for the [imath]y\text{-coordinate}[/imath] of the point [imath]\text{C} \ \text{or} \ \text{C}_y[/imath].

 

[Masaru]

Regarding:
The coordinates of point D = (700cos(220-90), 700sin(220-90))=(-449.9513268..., -536.2311102...)

Sorry, but I made an error when I was typing.
It should be:
The coordinates of point D = (700cos(90-220), 700sin(90-220))=(-449.9513268..., -536.2311102...)

Now I see where I went wrong.
The coordinates of point D should be:
(700cos(90-200), 700sin(90-200)) = (- 239.4141003...., -657.7848346...)

So,
DE = 992.820323 +239.4141003 = 1232.234423...
CE = 657.7848346 - 119.6152423 = 538.1695923...

So,
The angle CDE = tan-1(538.1695923.../1232.234423...) = 23.59295632....

Therefore, the bearing = 90 - 23.59295632.... = 66.40704368 = approximately 066 degrees as required.

 

[Masaru]

I would have found coordinates of B and C one at a time; no inverse tangent is needed in that step. But you have C correct. D is wrong due to the error I mentioned.

Then, to find the length and direction of DC, I would use the right triangle I've added here, which is just what you did:

View attachment 38394​

This is the same sort of triangle I'd use to find B and C.

I'm not sure any method could take less work than this. This is also essentially what you will do when you learn about vectors.

Yes, my apology for the error of the coordinates of point D.
The coordinates of point D should be:
(700cos(90-200), 700sin(90-200)) = (- 239.4141003...., -657.7848346...)
So the angle CDE = tan-1(538.17/1232.23) = approximately 23.59295632...
So the bearing = 90 - 23.59295632... = approximately 066 degrees.
Also thank you so much for drawing this comprehensive graph for me.