Answer
The boat must be pointed at an angle of $~~37^{\circ}~~$ west of north.
Work Step by Step
Let $v_y$ be the north component of the boat's velocity relative to the water.
Let $v_x$ be the west component of the boat's velocity relative to the water.
We can find an expression for $v_y$:
$v_x^2+v_y^2 = (4.0~m/s)^2$
$v_y^2 = 16-v_x^2$
$v_y = \sqrt{16-v_x^2}$
We can find an expression for the required net direction for the boat relative to the ground as an angle $\theta$ that is west of north:
$tan~\theta = \frac{82~m}{200`m}$
The net velocity of of the boat relative to the ground must also be in this direction $\theta$. We can find $v_x$:
$\frac{v_x - 1.1}{v_y} = tan~\theta = \frac{82~m}{200`m}$
$\frac{v_x - 1.1}{v_y} = 0.41$
$v_x - 1.1 = 0.41~v_y$
$v_x - 1.1 = 0.41~(\sqrt{16-v_x^2})$
$v_x^2 - 2.2~v_x+1.21 = 0.1681~(16-v_x^2)$
$1.1681~v_x^2 - 2.2~v_x-1.4796 = 0$
We can use the quadratic formula:
$v_x = \frac{2.2\pm \sqrt{(-2.2)^2-(4)(1.1681)(-1.4796)}}{(2)(1.1681)}$
$v_x = \frac{2.2\pm \sqrt{11.75}}{(2)(1.1681)}$
$v_x = -0.526~m/s, 2.41~m/s$
We can choose the positive solution.
We can find the required direction of the boat as an angle $\phi$ that is west of north:
$sin~\phi = \frac{2.41~m/s}{4.0~m/s}$
$\phi = sin^{-1}~(\frac{2.41~m/s}{4.0~m/s})$
$\phi = 37^{\circ}$
The boat must be pointed at an angle of $~~37^{\circ}~~$ west of north.
