I need help with the following trig problem and my tutor is out of town and out of touch...
A boat leaves a port and travels due east. From the port, there is an island 45.2 miles away on a bearing of 65.0º. After a boat has traveled 31.2 miles, find the bearing angle of the island from the boat. All bearing angles measures clockwise from due north.
here is what I have so far:
using the law of cosines, a^2=45.2^2 + 31.2^2- 2(45.2*31.2)cos(25), a=21.45
Then using the law of sines, I get (sinB/45.2)=(sin25/21.45)
SinB=.89055... Sin(^-1)(.89055)=B=62.9º
but B≠62.9º since it must be an obtuse angle in order to fit this application.
A=25º, a=21.45mi
B= 62.9º? b=45.2mi
C=92.1? c=31.2mi
Thanks in advance for your help. I know this is not really the place to look but I figured it is worth a shot.
A boat leaves a port and travels due east. From the port, there is an island 45.2 miles away on a bearing of 65.0º. After a boat has traveled 31.2 miles, find the bearing angle of the island from the boat. All bearing angles measures clockwise from due north.
here is what I have so far:
using the law of cosines, a^2=45.2^2 + 31.2^2- 2(45.2*31.2)cos(25), a=21.45
Then using the law of sines, I get (sinB/45.2)=(sin25/21.45)
SinB=.89055... Sin(^-1)(.89055)=B=62.9º
but B≠62.9º since it must be an obtuse angle in order to fit this application.
A=25º, a=21.45mi
B= 62.9º? b=45.2mi
C=92.1? c=31.2mi
Thanks in advance for your help. I know this is not really the place to look but I figured it is worth a shot.