Answer to Question #335167 in Analytic Geometry for math

Question

A plane with a velocity 400 km/h, 10 degrees, West of South encounters a wind with a

velocity of 40 km/h, from 35 degrees East of South. What is the plane’s resultant

velocity?

Answer 1

$$\vec{v}=-400\sin10\degree \vec{i}-400\cos10\degree\vec{j}$$

$$\vec{u}=40\sin35\degree \vec{i}-40\cos35\degree\vec{j}$$

$$\vec{v}_{res}=\vec{v}+\vec{u}$$

$$=(-400\sin10\degree +40\sin35)\vec{i}$$

$$+(-400\cos10\degree-40\cos35\degree)\vec{j}$$

$$\approx-46.5162\vec{i}-426.6892\vec{j}$$

$$|\vec{v}_{res}|=\sqrt{(-46.5162)^2+(-426.6892)^2}$$

$$\approx429.2172(km/h)$$

$$\tan \theta=\dfrac{-426.6892}{-46.5162}$$

$$\theta=264\degree$$

429.217 km/h, 6 degrees, West of South