Hello, repete!

$\displaystyle \text{Two forces }P\text{ and }Q\text{ with respective magnitudes 100N and 200N}$

. . $\displaystyle \text{are applied to a point.}$

$\displaystyle \text{The sum of the two forces is a horizontal force with magnitude 250N.}$

$\displaystyle \text}Find the angles that }P\text{ and }Q\text{ make with their sum.}$

$\displaystyle \text{(In the diagram, force }P\text{ is over and force }Q\text{ is under}$

. . $\displaystyle \text{the resultant horizontal force.)}$

Code:

P
o
* *
100 * * 200
* *
* *
* p *
O o - - - - - - - - - - - - - - o R
* q 250 *
* *
* *
200 * * 100
* *
o
Q

We have: .$\displaystyle \begin{array}{cccccccccc}

|\overrightarrow{OP}| \;=\;100 && p \:=\:\angle POR \\ \\[-4mm]

|\overrightarrow{OQ}| \;=\;200 && q \;=\;\angle QOR \\ \\[-4mm]

|\overrightarrow{OR}| \;=\;250 \end{array}$

$\displaystyle \text{In }\Delta OPR\!:\;\; \cos p \:=\:\dfrac{100^2 + 250^2 - 200^2}{2(100)(250)} \;=\;

0.65$

Hence: .$\displaystyle \angle p \;\approx\;49.46^o$

$\displaystyle \text{In }\Delta OQR\!:\;\;\cos q \:=\:\dfrac{200^2 + 250^2 - 100^2}{2(200)(250)} \;=\;0.925$

Hence:. . $\displaystyle \anglen q \;\approx\;22.33^o$