- #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.