+0

# Ship A is 60miles from a lighthouse on shore. Bearing measured from the lighthouse is S37deg.East. Ship B is 74 miles....

0
68
4
+136

Ship A is 60miles from a lighthouse on shore. Bearing measured from the lighthouse is S37deg.East. Ship B is 74 miles from the same lighthouse with a bearing of S48W measured from the house. Find the # of miles between the ships.

I tried to solve it. If it is incorrect then please if possible briefly tell me the steps on how to do it right? Regardless, thank you for whoever that takes out the time to check this!

Dec 11, 2019

#1
+107087
+3

You have uses Pythagoras's theorem.

This only works if the angle is 90 degrees,  Your angle is 48+37

You have to use the cosine rule to solve this.

Here is a tutorial on it if you need one.

https://www.mathsisfun.com/algebra/trig-cosine-law.html

Dec 11, 2019
#2
+136
+2

I did this:

= 60^2+74^2-2(60)(74)cos(85)

=3600+5476-8880*0.087155743

=8302.057002

sqrt of 8302.057002  = 91.11 miles

Is this right?

And thank you for your answer! It helped me understand it better

Dec 11, 2019
#3
+107087
+3

Yes that is good Roxettna except you presentation is just a little bit off.

AB^2   = 60^2+74^2-2(60)(74)cos(85)

AB^2  =3600+5476-8880*0.087155743

AB^2  =8302.057002

sqrt of 8302.057002  = 91.11 miles

so

AB = 91.11

The distance between the ships is approx 91 miles.

Since the other two distances are only written to the nearest mile, it makes sense that the third one would be good to the nearest whole mile as well.

But that does not matter.  91.11 miles is a good answer.

Melody  Dec 11, 2019
#4
+136
+2

Yes! Thank you very much for the clear instructions and everything! I'm so happy I got it right, minus some small exceptions

Roxettna  Dec 11, 2019