Bearing problems

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May 11th 2009, 08:30 PM
karlalujah

Bearing problems
I'm really having a hard time solving bearing problems. The only thing I can do is draw the triangles formed then I don't know what comes after. Can anyone help me with these?
- A ship is sailing due south. At 2:00 pm, an island is due S48°30’E of the ship. At 4:00 pm, the same island is due N32°20’E of the ship. Find
  - the distance of the ship from the island at 2 pm.
  - the time when the island is due east of the ship.
- A man left his house jogging at 6 am due north at the speed of 10 kph. At 7:00 he decided to change course and jogged due N68°W at the speed of 15 kph. Find
  - his distance from his house at 8:00 am
  - the bearing of his house from his position at 8:00 am.
- Two ships left a pier at 2:00 pm at the same time. The first ship is due S32°W sailing at the speed of 200 kph, while the other ship headed in the direction S25°E sailing at the speed of 150 kph. At 3:30 pm, the first ship stopped due to engine failure and its captain radioed the second ship for help. At that time instant, the captain of the second ship determined the bearing of the first ship and sailed towards it for help. If the second ship reached the first ship at 6:00 pm, at what speed did the second ship sailed towards the first ship? What was the bearing of the first ship from the second ship at 3:30 pm?
- A man left his house driving a car. He traveled in the direction N32°40’35” for 25 km, then changed course and traveled due west at the same speed for 20 km before stopping to rest. At that time instant, how far is the man from his house? What is his bearing from his house at that instant?
- A ship left a port at 12 noon sailed in the direction S32°15’30” at the speed of 120 kph. At 3:00 pm, the ship changed course and sailed in the direction S54°30’45” at the speed of 150 kph. Find the distance and bearing of the ship from the port at 6:00 pm.
May 12th 2009, 05:54 AM
Grandad
Hello karlalujah

Welcome to Math Help Forum!
Quote:
Originally Posted by karlalujah

I'm really having a hard time solving bearing problems. The only thing I can do is draw the triangles formed then I don't know what comes after. Can anyone help me with these?

A ship is sailing due south. At 2:00 pm, an island is due S48°30’E of the ship. At 4:00 pm, the same island is due N32°20’E of the ship. Find

the distance of the ship from the island at 2 pm.
the time when the island is due east of the ship.
You need to be told either the distance that the ship has sailed or its speed. Then use the Sine Rule to find the distance from the island at 2 pm.
Quote:
A man left his house jogging at 6 am due north at the speed of 10 kph. At 7:00 he decided to change course and jogged due N68°W at the speed of 15 kph. Find

his distance from his house at 8:00 am
the bearing of his house from his position at 8:00 am.
Use the Cosine Rule to find the distance, $d$ :

$d^2 = 10^2 +15^2 - 2.10.15.\cos 112^o$

Then use the Sine Rule to find another angle in the triangle, and hence the bearing.
Quote:
Two ships left a pier at 2:00 pm at the same time. The first ship is due S32°W sailing at the speed of 200 kph, while the other ship headed in the direction S25°E sailing at the speed of 150 kph. At 3:30 pm, the first ship stopped due to engine failure and its captain radioed the second ship for help. At that time instant, the captain of the second ship determined the bearing of the first ship and sailed towards it for help. If the second ship reached the first ship at 6:00 pm, at what speed did the second ship sailed towards the first ship? What was the bearing of the first ship from the second ship at 3:30 pm?
What are these - rocket ships? 200 and 150 kph? Anyway, again you'll have to used the Cosine Rule to find the distance apart of the ships at 3.30 pm. Then use speed = distance / time to find the speed of the second ship. Then use Sine Rule to find another angle and hence the bearing.
- Quote:
  
  A man left his house driving a car. He traveled in the direction N32°40’35” for 25 km, then changed course and traveled due west at the same speed for 20 km before stopping to rest. At that time instant, how far is the man from his house? What is his bearing from his house at that instant?
Same again here: Cosine Rule and then Sine Rule
- Quote:
  
  A ship left a port at 12 noon sailed in the direction S32°15’30” at the speed of 120 kph. At 3:00 pm, the ship changed course and sailed in the direction S54°30’45” at the speed of 150 kph. Find the distance and bearing of the ship from the port at 6:00 pm.
... and again here.

Here are some examples of the Sine Rule and the Cosine Rule if you're not sure how they work.

Grandad
May 12th 2009, 01:09 PM
aidan

Quote:

Originally Posted by karlalujah

I'm really having a hard time solving bearing problems. The only thing I can do is draw the triangles formed then I don't know what comes after. Can anyone help me with these?
...

Drawing the triangles is a good step.
Label all that you know or all that is given about the problem.
Then identify exactly what is wanted.
Once those pieces are set up, in general it becomes easy to see a method to solve the problem. Normally the teacher will supply a set of formulas that will be applicable to the problems given. Take a good look at your class notes.

By the way, in the first question, Grandad is right, there appears
to be a piece of information missing.

Distance typically is understood to mean a dimension in kilometers, miles, feet, meter, inches, etc.
However, it is possible that you are required to determine the time to get to the island from the 2pm position.
That is solvalble.

Quote:

A ship is sailing due south. At 2:00 pm, an island is due S48°30’E of the ship. At 4:00 pm, the same island is due N32°20’E of the ship. Find
the distance of the ship from the island at 2 pm.

The third angle, the angle at the island from the 2pm position to the 4pm position of the ship is:

180degress - 48deg30min - 32deg20min = 99deg10min
Then using the sine law:
$\frac {c}{\sin C} = \frac {a}{\sin A}$

$2 HOURS \times \frac {\sin 32deg20min}{\sin 99deg10min}$

The ship is 1.083 Hours from the island at 2pm
May 14th 2009, 01:30 AM
karlalujah

Thanks to you both! I'll take note of that. The 1st problem had missing data. I asked. I didn't get to answer these, though. Anyway, I'll just use the info for our exam. Thanks again!
September 11th 2009, 09:16 AM
ramiee2010

testing
$<br /> \frac{1}{x+iy}+\frac{1}{1+2i}=1\n<br /> \frac{1}{x+iy}=1-\frac{1}{1+2i}<br />$