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Question Number 208069 by Davidtim last updated on 03/Jun/24

$$F_1=4N,F_2=5N,R=8.32N$$$$\theta =?$$

Commented by mr W last updated on 03/Jun/24

$$\theta =180°-{\cos }^{-1}\frac{4^2+5^2-8.{32}^2}{2\times 4\times 5}=42.125°$$

Answered by efronzo1 last updated on 04/Jun/24

$$\mathrm{R}=\sqrt{4^2+5^2+40\cos \theta }$$$${(8.32)}^2-41=40\cos \theta$$$$\theta =\arccos \left(\frac{{(8.32)}^2-41}{40}\right)=45.125°$$

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