Thread: bearings

An attendant in a lighthouse receives a request for aid from a stalled craft located 5 miles due east of the lighthouse. The attendant contacts a second boat located 14 miles from the lighthouse on a bearing of N23degreesW. What is the distance and the bearing of the rescue boat from the stalled craft. without using law of cosine


sine23=x/14miles
x=14xsine23
x=5.47miles


cos23=a/14miles
a=14xcos23degrees
a=12.89miles


tan=sin/cos

You're on your way to the soulution.

So, as your intention,
tan(theta) = a / (x+5)
tan(theta) = 12.89 / (5.48 +5) = 1.23
theta = arctan(1.23) = 50.89 degrees
Or, maybe if you are looking for theta to the nearest drgree,
theta = 51 degrees

Hence, the bearing of the rescue boat from the stalled craft is
N 39deg W -------------answer.

The distance?

By Pythagorean theorem,
d = sqrt[(12.89)^2 +(5.48 +5)^2] = 16.61 miles. -----answer.

By trig,
sin(theta) = a / d
d = 12.89 / sin(50.89deg) = 16.61 miles

Or,
cos(theta) = (x +5) / d
d = (5.48 +5) / cos(50.89deg) = 16.61 miles.

thanks for the help!!