Let two vectors A and B makes an angle θ between them. The sum of the vectors or the magnitude of the resultant vector is given as,
![R=\sqrt[]{A^2+B^2+2AB\cos \theta}](https://img.qammunity.org/2023/formulas/physics/college/p9yv3cn0vcn295cj6gehq5gy50m1h498dt.png)
And the direction is given as,

Assuming east as positive x direction and north as positive y direction:
Given that,
Velocity of the plane;

Velocity of the wind;

The angle between North and East direction is,

The resultant velocity is given as,
![\begin{gathered} v_r=\sqrt[]{v^2_p+v^2_w+2v_pv_w\cos\theta} \\ =\sqrt[]{\lbrack(90)\rbrack^2+\lbrack(30)\rbrack^2+2\lbrack(90)\rbrack\lbrack(30)\rbrack^{}\cos (90\degree)}\text{ m/s} \\ \approx94.87\text{ m/s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/fom7luilbrou8flevekx5rtujo2f66cmuc.png)
The direction is given as,

Therefore, the resultant velocity of the plane is 94.87 m/s and is directed 18.43° towards North-East.