The Captain of a freighter 8km fromthe nearer of two unloading docks on the shore finds that the angle between the lines of sight to the docks is 35 degrees. if the docks are 10km apart, how far is the tanker from the farther dock?
The Captain of a freighter 8km fromthe nearer of two unloading docks on the shore finds that the angle between the lines of sight to the docks is 35 degrees. if the docks are 10km apart, how far is the tanker from the farther dock?
My answers :
A= unknown
B= unknown
C=35 degrees
a=unkown
b=8km
c=10km
using law of sin
sin 35/10=sinB/8
=.45
B=.45 dgrees
drwls, Saturday, May 24, 2008 at 5:19pm
The law of sines is the way to do this. That is your method #2. Method #1 only applies to right triangles.
You made some algebra errors, however.
sin B = 0.4589
B = 27.3 degrees
180 - B - C = A , not what you wrote.
i didn't get how he got 27 for B can someone explain it please
sin B = 0.8*sin 35
I used the law of sines, which you wrote correctly but did not calculate correctly, to get sin B. Then I took the arcsin of that to get the angle B itself