# Vectors

posted by .

I hope that the diagram I have drawn for myself matches the one in your question.

I have a triangle with sides 20 and 10 and an angle of 120º between them
By the cosine law I found the third side to be 26.4575.
You said you found the maginitude, I hope we match.

After that, use the Sine Law to find the angle across from the 10 side.

I got an angle of 9.7º
Subract that from 60º

A straight river flows east at a speed of 10mi/h. A boater starts at the south shore of the river and heads in a direction 60degree from the shore. Diagram is given. The motorboat has a speed of 20mi/h relative to the water.
Find the true speed and direction of the motorboat.
I know how to find the true speed which is the magnitude. But which angle is the direction? Since one angle is given 60degree, can't the other be 30degree? Please help.

Magnitude is right. 26.4575
But the direction is 49.1 degrees.

So, it is not a right triangle?

no it is not a right-angled triangle.

I drew my river going horiontally and considered the x-axis the south side of the river.
I drew a line coming up 60º from the x-axis and labelled it 20 units long.
I then drew a horizontal line, the river flow, from the end of my first line and made that 10 units long.
Now join that endpoint to the origin.
This line is the resultant vertor and it s length is the maginitude which we found to be 26.45...

so far ok?

by Sine Law:
sin(theta)/10 = sin120 /26.45..
sin(theta) = .3273..
so theta = 19.1 (that is the angle across the side of 10)

so the angle of the resultant up from the x-axis is 60-19.1 = 40.9

but using navigation directions, which measure from the North and going clockwise, we would have 90-40.9