Asked by Will
A boat is going 30 mph. It's direction is 100 degrees east of north. The water current is 15 mph and has a direction of 22 degrees east of north. Find the magnitude, and direction of boat.( with water current)
Answers
Answered by
Reiny
ok, you fixed your earlier typo.
Did you make a sketch?
I completed the parallogram and am looking at a triangle with sides 30 and 15 and the angle between them as 102°
by the cosine law:
R^2 = 30^2 + 15^2 - 2(30)(15)cos102
= 1312.102...
R = appr 36.22 mph
by sine law:
sinØ/15 = sin102/36.223..
sinØ = .40505..
Ø = 23.89°
so the direction is E 13.89° N
or in your notation: 76.1° east of north
alternate solution:
first vector = (30cos(-10) , 30sin(-10)
2nd vector = (15cos68 , 15sin68)
resultant = (30cos(-10) , 30sin(-10) + (15cos68 , 15sin68)
= (29.544 , -5.209) + (5.619 , 13.908)
= (35.163 , 8.698)
magnitude = √(35.163^2 + 8.698^2)
= 36.223 , same as before
for angle:
tan O = 8.698/35.163
angle O = 13.894° same as before
Did you make a sketch?
I completed the parallogram and am looking at a triangle with sides 30 and 15 and the angle between them as 102°
by the cosine law:
R^2 = 30^2 + 15^2 - 2(30)(15)cos102
= 1312.102...
R = appr 36.22 mph
by sine law:
sinØ/15 = sin102/36.223..
sinØ = .40505..
Ø = 23.89°
so the direction is E 13.89° N
or in your notation: 76.1° east of north
alternate solution:
first vector = (30cos(-10) , 30sin(-10)
2nd vector = (15cos68 , 15sin68)
resultant = (30cos(-10) , 30sin(-10) + (15cos68 , 15sin68)
= (29.544 , -5.209) + (5.619 , 13.908)
= (35.163 , 8.698)
magnitude = √(35.163^2 + 8.698^2)
= 36.223 , same as before
for angle:
tan O = 8.698/35.163
angle O = 13.894° same as before
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