# Thread: Finding degree and bearing

1. ## Finding degree and bearing

Town C is 5 miles due east of town D.
Town E is 12 miles from town C at a bearing (from C) of N52E.

Two parts of the problem, I'm not sure about the second part:

1: How far apart are towns DE?

Using the law of cosines, I found that DE is 16 miles.

2: Find the bearing of two E from town D, round to nearest one-half mile.

Using the law of sines, I found that angle D is 11 degrees. If I use the NS line, and assume that D is 11 degrees of a full 90 degree angle, I can subtract 11 from 90, resulting in 79 degrees.

Am I calculating part two correctly?

Thank you!

2. Originally Posted by Snowcrash
Town C is 5 miles due east of town D.
Town E is 12 miles from town C at a bearing (from C) of N52E.

Two parts of the problem, I'm not sure about the second part:

1: How far apart are towns DE?

Using the law of cosines, I found that DE is 16 miles.

2: Find the bearing of two E from town D, round to nearest one-half mile.

Using the law of sines, I found that angle D is 11 degrees. If I use the NS line, and assume that D is 11 degrees of a full 90 degree angle, I can subtract 11 from 90, resulting in 79 degrees.

Am I calculating part two correctly?

Thank you!
agree with #1 ... DE = 16.2 miles

using the law of sines, I get $\displaystyle m\angle{D} = \arcsin\left(\frac{12\sin(142^{\circ})}{DE}\right) \approx 27^{\circ}$

3. Thanks - I do see that angle E = 27 degrees - would the bearing from D to E then be S27W?

I thought when the problem asked the bearing from town D to E, I'd need to find the angle of Town D because it's "From D to E" and the result would be N79E for bearing. Setting up word problems is always difficult for me!

4. Originally Posted by Snowcrash
Thanks - I do see that angle E = 27 degrees - would the bearing from D to E then be S27W?

I thought when the problem asked the bearing from town D to E, I'd need to find the angle of Town D because it's "From D to E" and the result would be N79E for bearing. Setting up word problems is always difficult for me!
$\displaystyle m\angle{E} = 11^{\circ}$

bearing of town E from town D would be $\displaystyle N 63^{\circ} E$

look at the rough layout ...

5. Eek - I'm still not seeing it - sorry - I do have my drawing as yours is set up too -

So, to find angle E we use the law of sines I did this:

sin E / 5 = sin 142 / 16
16 sin E = 5 sin142
sin E = 5 sin142 / 16

As I'm typing this I think I see what I did - I flipped the values, using the side length 5 instead of 12. I came up with the same answers, just reversed.

Thanks Skeeter!

6. Originally Posted by Snowcrash
Eek - I'm still not seeing it - sorry - I do have my drawing as yours is set up too -

So, to find angle E we use the law of sines I did this:

sin E / 5 = sin 142 / 16
16 sin E = 5 sin142
sin E = 5 sin142 / 16

$\displaystyle \textcolor{red}{m\angle{E} = \arcsin\left(\frac{5\sin(142)}{16.2}\right) = 11}$ degrees
...