Question Number 208069 by Davidtim last updated on 03/Jun/24
$${F}_{\mathrm{1}} =\mathrm{4}{N},\:\:\:{F}_{\mathrm{2}} =\mathrm{5}{N},\:\:\:{R}=\mathrm{8}.\mathrm{32}{N} \\ $$$$\theta=? \\ $$
Commented by mr W last updated on 03/Jun/24
$$\theta=\mathrm{180}°−\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{8}.\mathrm{32}^{\mathrm{2}} }{\mathrm{2}×\mathrm{4}×\mathrm{5}}=\mathrm{42}.\mathrm{125}° \\ $$
Answered by efronzo1 last updated on 04/Jun/24
$$\:\mathrm{R}=\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} +\mathrm{40}\:\mathrm{cos}\:\theta} \\ $$$$\:\left(\mathrm{8}.\mathrm{32}\right)^{\mathrm{2}} −\mathrm{41}=\:\mathrm{40}\:\mathrm{cos}\:\theta \\ $$$$\:\:\theta=\:\mathrm{arccos}\:\left(\frac{\left(\mathrm{8}.\mathrm{32}\right)^{\mathrm{2}} −\mathrm{41}}{\mathrm{40}}\right)=\mathrm{45}.\mathrm{125}° \\ $$