Question 803720
You may be able to draw this picture:

Horizontal axis, current of river toward the right; vertical axis, upward is "accross" the river straight up.


The boat is trying to go into quadrant 4, so that direction of the boat vector is 140 degrees, using  "to the right..." as the reference direction, the positive x axis.  The river vector is just toward the right, no angle, since angle 0 is unrotated and toward the right.


b for the boat's magnitude, r for the river's magnitude.
The boat vector is {{{b*cos(140)+b*sin(140)}}}
The river vector is {{{r*cos(0)+r*sin(0)}}}
The sum of the vectors for b and r is {{{10*(cos(90)+sin(90))}}}


Adding the components will give you two equations.
{{{b*cos(140)+r*cos(0)=10*cos(90)}}}  and {{{b*sin(140)+r*sin(0)=10*sin(90)}}}
{{{b(-0.766)+r=10*0}}}  and  {{{b*(0.643)+r*0=10*1}}}
{{{-0.766b+r=0}}} and {{{0.643b=10}}}
'
{{{r=0.766b}}} and {{{b=10/(0.643)}}}


Second part simplified is then {{{highlight(b=15.55)}}}, and therefore {{{r=0.766*15.55=highlight(11.91)}}}.
If you prefer for proper accuracy, {{{b=16}}} and {{{r=12}}}.