as far as I can make it, the vectors will be like the ones in the picture below, where the green one is the motorboat and the current is in red, and their sum will be the gray one.
since the boat is going 6 m/s in 1 second the motorboat has a magnitude of 6, and since it's due East, is just a horizontal line with an angle of 0°.
the current is due North, therefore has an angle of 90° and since the current is going 3 m/s, so in 1 second it has a magnitude of 3.
[tex] \bf \stackrel{\textit{motorboat}}{\begin{cases}x=rcos(\theta )\\\qquad 6cos(0^o)\\\qquad 6\\y=rsin(\theta )\\\qquad 6sin(0^o)\\\qquad 0\end{cases}}\implies <6,0>\qquad \stackrel{\textit{current}}{\begin{cases}x=3cos(90^o)\\\qquad 0\\y=3sin(90^o)\\\qquad 3\end{cases}}\implies <0,3>\\\\\\<6,0>+<0,3>\implies <6+0,0+3>\implies <\stackrel{a}{6},\stackrel{b}{3}>\\\\\\\measuredangle \theta =tan^{-1}\left(\cfrac{3}{6} \right)\implies \measuredangle \theta \approx 26.57^o [/tex]
so, that will be the angle of the resultant vector, now, we know the river is 100 meters wide, so we can use cosine of θ to get the length and therefore the magnitude of the resultant vector. Keeping in mind that the angle of elevation θ is about 26.57° and that the adjacent side is 100.
[tex] \bf cos(26.57^o)=\cfrac{\stackrel{adjacent}{100}}{\stackrel{hypotenuse}{h}}\implies h=\cfrac{100}{cos(26.57^o)}\implies h\approx 111.8 [/tex]
