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**Hint:-**The vector resultant of the wind directions can be used to find the direction in which the pilot should head also the magnitude of the resultant can be used to find the speed of the plane with respect to the ground.

Formula used: If there are two velocity vectors ${\vec V_A}$ and ${\vec V_B}$ then the velocity vector of ${\vec V_A}$ with respect to ${\vec V_B}$ is given by,

${\vec V_{\dfrac{A}{B}}} = {\vec V_A} - {\vec V_B}$

**Complete step-by-step solution**

It is given in the problem that the aeroplane pilot wishes to fly the plane in the west direction and the wind is flowing in the west direction. The velocity of the aeroplane in still air is $300km{h^{ - 1}}$ and the speed of the wind is$100km{h^{ - 1}}$.

The angle between the speed of wind and the speed of the plane w.r.t plane is given by$\theta $.

The velocity of the plane w.r.t air is equal to ${\vec V_{\dfrac{p}{A}}} = 300km{h^{ - 1}}$and the velocity of air is given by ${\vec V_A} = 100km{h^{ - 1}}$.

The velocity of the place with respect air is given by,

$ \Rightarrow {\vec V_{\dfrac{p}{A}}} = {\vec V_p} - {\vec V_A}$

$ \Rightarrow {\vec V_p} = {\vec V_{\dfrac{p}{A}}} + {\vec V_A}$.

Let us calculate the value of $\sin \theta $ from the diagram.

$ \Rightarrow \sin \theta = \dfrac{{{{\vec V}_A}}}{{{{\vec V}_{\dfrac{p}{A}}}}}$

$ \Rightarrow \sin \theta = \dfrac{{100}}{{300}}$

$ \Rightarrow \sin \theta = \dfrac{1}{3}$

$ \Rightarrow \theta = {\sin ^{ - 1}}\dfrac{1}{3}$

The direction in which the plane should head is equal to $\theta = {\sin ^{ - 1}}\dfrac{1}{3}$.now let us calculate the speed of the plane with respect to ground.

The speed of plane with respect to ground is equal to,

$ \Rightarrow {V_p} = {V_{\dfrac{p}{A}}} \cdot \left( {\cos \theta } \right)$

$ \Rightarrow {V_p} = \left( {300} \right) \cdot \left( {\sqrt {1 - {{\sin }^2}\theta } } \right)$

$ \Rightarrow {V_p} = \left( {300} \right) \cdot \left( {\sqrt {1 - {{\sin }^2}\theta } } \right)$

Since, $\sin \theta = \dfrac{1}{3}$.

$ \Rightarrow {V_p} = \left( {300} \right) \cdot \left( {\sqrt {1 - {{\left( {\dfrac{1}{3}} \right)}^2}} } \right)$

$ \Rightarrow {V_p} = \left( {300} \right) \cdot \left( {\sqrt {\dfrac{8}{9}} } \right)$

$ \Rightarrow {V_p} = \dfrac{{300 \times 2\sqrt 2 }}{3}$

$ \Rightarrow {V_p} = 200\sqrt 2 km{h^{ - 1}}$

The velocity of the plane with respect to ground is equal to ${V_p} = 200\sqrt 2 km{h^{ - 1}}$.

**Note:-**The resultant direction of the velocity of wind and plane should be the direction in which the pilot should fly the plane. The use of trigonometry in geometry is helping us to solve and get the resultant direction of the velocities of wind and plane.

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