#### Known

The relative motion of the boat with respect to the ground is equal to:

$$
begin{align*}
boxed{vec{v}_{tx{bg}}=vec{v}_{tx{bw}}+vec{v}_{tx{wg}}}
end{align*}
$$

#### Calculation

Given: $v_{tx{bw}}=6.1 frac{tx{m}}{tx{s}}$ with a direction relative to water $theta=25^circ$, $v_{tx{wg}}=4.5 frac{tx{m}}{tx{s}}$, working with the components of $vec{v}_{tx{bw}}$ and $vec{v}_{tx{wg}}$, we have:

$$
begin{align*}
&v_{tx{bw}, x}=left(6.1 frac{tx{m}}{tx{s}}right) tx{cos}(25^circ)=5.5 frac{tx{m}}{tx{s}}\
&v_{tx{bw}, y}=left(6.1 frac{tx{m}}{tx{s}}right) tx{sin}(25^circ)=2.6 frac{tx{m}}{tx{s}}\
&v_{tx{wg}, x}=0 frac{tx{m}}{tx{s}}\
&v_{tx{wg}, y}=-4.5 frac{tx{m}}{tx{s}}\
end{align*}
$$

Therefore:

$$
begin{align*}
&v_{tx{bg}, x}=v_{tx{bw}, x}+v_{tx{wg}, x}=5.5 frac{tx{m}}{tx{s}}+0 frac{tx{m}}{tx{s}}=5.5 frac{tx{m}}{tx{s}}\
&v_{tx{bg}, y}=v_{tx{bw}, y}+v_{tx{wg}, y}=2.6 frac{tx{m}}{tx{s}}-4.5 frac{tx{m}}{tx{s}}=-1.9 frac{tx{m}}{tx{s}}
end{align*}
$$

This implies:

$$
begin{align*}
v_{tx{bg}}&=sqrt{left(v_{tx{bg}, x}right)^2+left(v_{tx{bg}, y}right)^2}\
&=sqrt{left(5.5 frac{tx{m}}{tx{s}}right)^2+left(-1.9 frac{tx{m}}{tx{s}}right)^2}\
&=5.8 frac{tx{m}}{tx{s}}
end{align*}
$$

And

$$
begin{align*}
theta=tx{tan}^{-1}left(frac{-1.9 cancel{frac{tx{m}}{tx{s}}}}{5.5 cancel{frac{tx{m}}{tx{s}}}}right)=-19^circ tx{or} 19^circ tx{downstream}
end{align*}
$$

—

#### Conclusion

The boat is moving with a speed of $5.8 frac{tx{m}}{tx{s}}$ and $19^circ$ downstream relative to the ground.

Graphically:

#### Known

The relative motion of the airplane with respect to the ground is equal to:

$$
begin{align}
boxed{vec{v}_{tx{pg}}=vec{v}_{tx{pa}}+vec{v}_{tx{ag}}}
end{align}
$$

Where $vec{v}_{tx{pg}}$ is the velocity of the plane with respect to the ground, $vec{v}_{tx{pa}}$ is the velocity of the plane with respect to the air and $vec{v}_{tx{ag}}$ is the velocity of the air with respect to the ground (wind velocity).
#### Calculation

Givens: $v_{tx{ag}}=65 frac{tx{km}}{tx{h}}$, from east to west. $v_{tx{pa}}=340 frac{tx{km}}{tx{h}}$.

The pilot wants to travel north, so the plane will have $v_{tx{pg}, y}$ component on the $y$ axis and $v_{tx{pg}, x}=0 frac{tx{km}}{tx{h}}$ component on the $x$ axis. See Figure 1.

The components of the wind velocity are given by:

$$
begin{align*}
v_{tx{ag}, x}=-65 frac{tx{km}}{tx{h}}, v_{tx{ag}, y}=0 frac{tx{km}}{tx{h}}
end{align*}
$$

Choosing the positive direction from west to east and the negative from east to west.

Finally, the components of the plane’s velocity with respect to the air are:

$$
begin{align*}
v_{tx{pa}, x}=v_{tx{pa}} tx{cos}(theta)hspace{0.5cm} tx{and}hspace{0.5cm} v_{tx{pa}, y}=v_{tx{pa}} tx{sin}(theta)
end{align*}
$$

Using expression (1) in components:

$$
begin{align*}
v_{tx{pg}, y}=v_{tx{pa}, y}+v_{tx{ag}, y}
end{align*}
$$

And

$$
begin{align*}
&v_{tx{pg}, x}=v_{tx{pa}, x}+v_{tx{ag}, x}\
&v_{tx{pg}, x}=v_{tx{pa}} tx{cos}(theta)+v_{tx{ag}, x}\
&implies tx{cos}=frac{v_{tx{pg}, x}-v_{tx{ag}, x}}{v_{pa}}=frac{0 frac{tx{km}}{tx{h}}-left(-65 frac{tx{km}}{tx{h}}right)}{340 frac{tx{km}}{tx{h}}}=frac{65}{340}\
&thereforetheta=tx{cos}^{-1}left(frac{65}{340}right)=79^circ
end{align*}
$$

—

#### Conclusion

The pilot should steer the plane $79^circ$ north of east.

#### Known

The relative motion of the boat with respect to the ground is given by:

$$
begin{align*}
&vec{v}_{tx{bg}}=vec{v}_{tx{bw}}+vec{v}_{tx{wg}}\
implies &vec{v}_{tx{bg}, x}=vec{v}_{tx{bw}, x}+vec{v}_{tx{wg}, x}\
&vec{v}_{tx{bg}, y}=vec{v}_{tx{bw}, y}+vec{v}_{tx{wg}, y}
end{align*}
$$