- #1

rannasquaer

- 7

- 0

Given the triangle above where \(\displaystyle V < v'_{1}\), prove that the \[ v_{1}=V \cos(\psi)+v'_{1} \cos(\theta - \psi) \]

It is said that \(\displaystyle v_{1}\) is equal to the sum of the orthogonal projections on \(\displaystyle v_{1}\) of \(\displaystyle V\) and of \(\displaystyle v'_{1}\) and that is precisely the expression that show. But I couldn't see how to make the projection and the calculations.