Perpendicular forces

Distribute the force of magnitude F = 100 N into two perpendicular components with magnitudes F1 and F2 so that the angle between forces F1 and F is 43°52'.

Correct answer:

F₁ =  72.0954 N
F₂ =  69.2983 N

Step-by-step explanation:

$$\begin{gathered} F=100\text{N} \\ \alpha =43°52^{\prime }=43°+\frac{52}{60}°=43.8667°\doteq 43.8667 \\ \cos \alpha =F_1:F \\ {F}_1=F\cdot \cos \alpha =F\cdot \cos 43.866666666667°=100\cdot \cos 43.866666666667°=100\cdot 0.720954=72.095=72.0954\text{N} \end{gathered}$$

$$\begin{gathered} \sin \alpha =F_2:F \\ {F}_2=F\cdot \sin \alpha =F\cdot \sin 43.866666666667°=100\cdot \sin 43.866666666667°=100\cdot 0.692983=69.298=69.2983\text{N} \end{gathered}$$

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