# Thread: Trigonometric help in mechanics question?

1. ## Trigonometric help in mechanics question?

While solving a projectile motion problem, i found the elevation angle as
θ = arccos ((2/3) + (v^2/3gR)) (which is correct according the solution). The next step requires tan θ, which I calculated as equal to √(5 - (v^4/g^2.R^2) - (4v^2/gR))/ (2 + (v^2/gR)). The solution in the text says that tan θ = - √(1 - (v^2/gR))/(√(2 + (v^2/gR))). Can anyone help me on what I am missing?

2. Hello, fardeen_gen!

While solving a projectile motion problem, i found the elevation angle as:
$\theta \:=\:\arccos\left(\tfrac{2}{3} + \tfrac{v^2}{3gR}\right)$, (which is correct according the solution).

The next step requires $\tan\theta$, which I calculated as equal to:
. . $\frac{\sqrt{5 - \frac{v^4}{g^2R^2} - \frac{4v^2}{gR}}} {2 + \frac{v^2}{gR}}$
The text says that: . $\tan\theta \:=\: -\frac{\sqrt{1 - \frac{v^2}{gR}}} {\sqrt{2 + \frac{v^2}{gR}}}$

We have: . $\theta \:=\:\arccos\left(\frac{2gR+v^2}{3gR}\right) \quad\Rightarrow\quad\cos\theta \:=\:\frac{2gR + v^2}{3gR} \:=\:\frac{adj}{hyp}$

Hence, $\theta$ is in a right triangle with: . $adj = 2gR + v^2,\;hyp = 3gR$

. . Using Pythagorus, we have: . $opp \,=\,\sqrt{5g^2R^2 - 4gRv^2 - v^4}$

Therefore: . $\tan\theta \;=\;\frac{opp}{adj} \;=\;\frac{\sqrt{5g^2R^2 - 4gRv^2 - v^4}}{2gR + v^2}$