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