1. bearing problems

I'm really having trouble with bearing problems. Can someone please help me solve them? I just don't know how and when to use Cosine and Sine Law. This is the first time I encountered them.

1.) A plane flew 50km on a bearing 63°20', and then flew on a bearing of 153°20' for 140km. Find the distance between the starting point and the ending point.

2.) A boat left the port on a course of 230°20' for 1.5hour at 210mph and then at 140°20' Edit*-(for 2 hr at 180mph)-. How far is the boat from the port? In what direction must it travel to return to the port?

3.) City A and City B are located on a north-south line and 34.6 miles apart. An observer on the east of that line finds that the directions of A and B are N27°10'W and S62°50'W, respectively. How far is he from each city?

4.) At a certain time, a lighthouse is south of a ship. Thirty minutes later, the ship bears N40°20'E from the lighthouse. If the ship is sailing east at 20kph, find the distance of the ship from the lighthouse at each time of observations.

5.) A ship leaves its home port and sails on a bearing of N38°15'E at 24mph. At the same instant. another ship on the same port on a bearing of S51°45'E at 28mph. Find the distance between the 2 ships after 4 hrs.

6.) A surveyor describes a tract of land by saying : starting at a concrete stake, thence N37°20'E for 120km, thence S52°40'E until due east of the starting point, thence to the starting point. Find the perimeter of the tract.

Any help would be greatly appreciated.

2. Originally Posted by angeIi28
I'm really having trouble with bearing problems. Can someone please help me solve them? I just don't know how and when to use Cosine and Sine Law. This is the first time I encountered them.

1.) A plane flew 50km on a bearing 63°20', and then flew on a bearing of 153°20' for 140km. Find the distance between the starting point and the ending point.

2.) A boat left the port on a course of 230°20' for 1.5hour at 210mph and then 140°20' at 180mph. How far is the boat from the port? In what direction must it travel to return to the port?

3.) City A and City B are located on a north-south line and 34.6 miles apart. An observer on the east of that line finds that the directions of A and B are N27°10'W and S62°50'W, respectively. How far is he from each city?

4.) At a certain time, a lighthouse is south of a ship. Thirty minutes later, the ship bears N40°20'E from the lighthouse. If the ship is sailing east at 20kph, find the distance of the ship from the lighthouse at each time of observations.

5.) A ship leaves its home port and sails on a bearing of N38°15'E at 24mph. At the same instant. another ship on the same port on a bearing of S51°45'E at 28mph. Find the distance between the 2 ships after 4 hrs.

6.) A surveyor describes a tract of land by saying : starting at a concrete stake, thence N37°20'E for 120km, thence S52°40'E until due east of the starting point, thence to the starting point. Find the perimeter of the tract.

Any help would be greatly appreciated.
1. You don't need any Sine or Cosine (except #2!) because all problems are about right triangles. Simple Pythagorean theorem will do.

2. The problem with your problems is that you have to convert the bearings and courses into a more convenient format of angle measure:
Courses are given wrt to North N = 0°, measured clockwise. Thus
N = 0°
E = 90°
S = 180°
W = 270°

A course of S62°50'W = 180° + 62°50' = 242°50'
A course of S52°40'E = 180° - 52°40' = 179°60' - 52°40' = 127°20'

3. Draw a rough sketch for each problem.

4. The problem #2 can't be solved because the traveling time of the 2nd course is missing.

3. Originally Posted by angeIi28
I'm really having trouble with bearing problems. Can someone please help me solve them? I just don't know how and when to use Cosine and Sine Law. This is the first time I encountered them.
...
Any help would be greatly appreciated.
As earboth stated, you don't need to use the trig laws,
but you may do so as as option.

For all of the calculations:
distance x sin(bearing) = y component
distance x cos(bearing) = x component

Problem 1):
50 x sin(63°20') = 50 x 0.89363 = 44.68 = y1
50 x cos(63°20') = 50 x 0.4488 = 22.44 = x1

140 x sin(153°20') = 140 x 0.4488 = 62.83 = y2
140 x cos(153°20') = 140 x -0.89363 = -125.11 = x2

$Y_{endingpoint} = y_1 + y_2 = 44.68+62.83 = 107.51$

$X_{endingpoint} = x_1 + x_2 = 22.44 + (-125.11) = -102.67$

The distance between the starting point and the ending point:

$\sqrt{ 107.51^2 + (-102.67)^2}$

Since the sin of one bearing equals the cos of the other bearing (except for sign) & the cos of the one bearing equals the sin of the other bearing, you have a right triangle.

$\sqrt{ 50^2 + 140^2}$

Both results should be equal.
You may need more decimal precision that I have given here.

NOTE:
When North is used the sin value will be positive.
When South is used the sin value will be NEGATIVE.
That means that the y component will be positive or negative.

When East is used the cos value will be positive.
When West is used the cos value will be negative.
That means that the x component will be positive or negative.

---
NOTE 2:
You can use the cosine law on this:

The difference between the two bearings (63°20' and 153°20') is angle C.

the distance between the starting & ending:

$c^2 = a^2 + b^2 - 2ab \, \cos C$
$c^2 = 50^2 + 140^2 - 2 \times 50 \times 140 \times \cos (153deg20min - 63deg20min)$

Then take the square root of c^2

4. thanks

thanks for the help... I still feel uneasy about bearing problems but now I'm beginning to understand it. Thanks again