A person starts at Camp 1 and walks 4km..... (one of these types of questions)

[ripple]

Hey guys,
not sure how to do these

1. A person starts at Camp 1 and walks 4km due South and then turns and walks for 15km on a bearing of N 20^◦ E to get to Camp 2. The person then decides to head straight back to Camp 1 without deviating.

(i) How far and on what bearing should the person walk to get back from Camp 2 to Camp 1?
(ii) After reaching Camp 2, how far East and North from Camp 1 is the person?

Our professor bombards us with a new topic each week. Week one Algebra, week 2 set theory, week 3 logarithms, week 4 trig etc. I thought it would be ok since we wouldn't really go into it that deep however no matter how much time I put in, I can never cover the given material for each subject each week!

Anyway, just wondering best way to approach these types of questions. Thanks in advance!

 

[Deleted member 4993]

Hey guys,
not sure how to do these

1. A person starts at Camp 1 and walks 4km due South and then turns and walks for 15km on a bearing of N 20^◦ E to get to Camp 2. The person then decides to head straight back to Camp 1 without deviating.

(i) How far and on what bearing should the person walk to get back from Camp 2 to Camp 1?
(ii) After reaching Camp 2, how far East and North from Camp 1 is the person?

Our professor bombards us with a new topic each week. Week one Algebra, week 2 set theory, week 3 logarithms, week 4 trig etc. I thought it would be ok since we wouldn't really go into it that deep however no matter how much time I put in, I can never cover the given material for each subject each week!

Anyway, just wondering best way to approach these types of questions. Thanks in advance!

Start by using a sketch.

Have you drawn a sketch of the problem?

The person's path (locus) will be a triangle. Have you drawn that triangle?

 

[Ron Spain]

Good advice to start with a drawing. Then I would imagine the North-South dimension is y and the East-West dimension is x, and calculate the coordinates of each point by adding the sine times the distance for x and cosine times the distance for y for each leg of the journey.

 

[ripple]

Start by using a sketch.

Have you drawn a sketch of the problem?

The person's path (locus) will be a triangle. Have you drawn that triangle?

yeah i've got the triangle drawn, just not sure how the whole bearings thing works

 

[HallsofIvy]

Drawing your picture with "north" at the top, "east" to the right, etc., as on a map, "bearing N 20 E" means a line 20 degrees to the right of north.

1. A person starts at Camp 1 and walks 4km due South and then turns and walks for 15km on a bearing of N 20◦ E to get to Camp 2. The person then decides to head straight back to Camp 1 without deviating.

So your first line is a vertical line, "Camp 1" at the top, length 4 km (whatever scale you choose to use- say 4 inches to represent 4 km) then draw a line at the bottom of that 20 degrees to the right and 15 cm long. Finally, draw a line from the end of that line back to "Camp 1". You will have a triangle with two sides of length 4 and 16 inches (km) with a 20 degree angle between them. You can use the sine law to determine the other two angles and the length of the third side- taking "A" to be the angle at "Camp 1", "B" the third angle (where the person turns to walk back to the camp, and "x" the length of the third side (the distance he finally walks to "Camp 1") we have, by the "sine law"
\(\displaystyle \frac{sin(A)}{15}= \frac{sin(B)}{4}= \frac{sin(20)}{x}\).

Of course, since the angles in any triangle add to 180 degrees, B= 180- A so we have the two equations \(\displaystyle \frac{sin(A)}{15}= \frac{sin(180- A)}{4}\) to solve for A, and \(\displaystyle \frac{sin(A)}{15}= \frac{sin(20)}{x}\) to then solve for x.