Vector and cosine law?

[saac]

Vector and cosine law??

Homework Statement

Three deer, A, B, and C, are grazing in a field. deer B is located 62m from deer A at an angle of 51 north of west. deer C is located 77 degree north of east relative to deer A. The distance between deer B and C is 95m. What is the distance between deer A and C?

Homework Equations

fairly new to physics and I am stuck on these types of problems. Not sure how to draw the graph or complete the calculations. Any tips would be appreciated

The Attempt at a Solution

 

[Simon Bridge]

Just draw the triangle.
Get some graph paper and select a scale (hint 1cm = 1m).
Pick a direction for north (hint: up the page)
[if north is the y axis, then south is -y, east is +x and west is -x.]

Mark point A someplace handy.
Use protractor to get 51 degrees upwards from the -x axis (N of W)
Draw a line - to scale - 62m long.
Mark the end of the line with a point, label it B.

Follow the instructions like that for the rest.

You can either measure the length needed with a ruler or use trigonometry.

The trig method either resolves each position vector wrt to the x and y-axis to get the points B and C, then compute |C-B|
OR use the cosine rule ... for which you need the angle between the two position vectors.
To do that you need to realize that the E-W line forms 180 degrees.
Your diagram will show you how this relates to the angles you are given.

 

[saac]

So i have drawn what i think is correct. Do i now separate the large triangle into two separate ones and then solve Hypotenuse for triangle A-B?

 

[NascentOxygen]

So i have drawn what i think is correct. Do i now separate the large triangle into two separate ones and then solve Hypotenuse for triangle A-B?

No. You have a triangle ABC where you know the lengths of two sides, and you know one angle. Using your knowledge of trigonometry, that is sufficient information to determine all angles and all sides.

Do you know the sine rule? If not, try a google search.

 

[Simon Bridge]

If you know two sides and the angle between them, that's the cosine rule.
The sine rule is where you know one side and two angles - one opposite the known side.
But there is more than enough information here that normal trigonometry is sufficient.

You have drawn the triangle ABC?
You have two angles about vertex A (at the origin)?

There are two methods - both equivalent.

1.0 normal trig
1.1 put point A at the origin and +y pointing N
1.2 find the coordinates of B and C (trigonometry)
1.3 compute a = |B-C| (pythagoras)

2.0 shortcut
1.1 put b=|AB|, c=|AC|, find θ = angle BAC
1.2 put these numbers into the cosine rule formula

 

[NascentOxygen]

2.0 shortcut
1.1 put b=|AB|, c=|AC|, find θ = angle BAC
1.2 put these numbers into the cosine rule formula

We don't know |AC|. So attempting to use the cosine rule yields a quadratic to be solved.

Not such a shortcut.

 

[Simon Bridge]

Oh you are correct, I misread.
In that case you do need the sine rule to find angle BCA - it is a Side-Side-Angle problem.
Well spotted.

 

[saac]

Thanks for the replies but I've had to move on as the class is quickly progressing