A boat in calm seas travels in a straight line and ends the trip 22 km west and 53 km north of its original position. To the nearest tenth of a degree, find the direction of the trip.
I did show the work......
tangent = opp/adjacent
= 53 / -22
arctan 53 / -22 = 112.5 degrees (it is in the second quadrant y is positive (53) x is negative (22)
+2862 SOH CAH TOA
We use TOA in this case because only the legs of a triangle is used. We are also calculating the angle of elevation.
Opposite = 53
Adjacent = 22
Tan(x) = 53/22
x = arctan(53/22)
Is this correct?
TBH web2.0calc should implement a drawing program where you can draw something quickly and insert it.
