Question-187716

Question Number 187716 by yaslm last updated on 20/Feb/23

Answered by Ar Brandon last updated on 20/Feb/23

ΣF_x : 300sinβ=200sinα ⇒sinβ=((2sinα)/3) ΣF_y : 300cosβ+200cosα=400 ⇒3(√(1−sin^2 β))+2cosα=4 ⇒3(√(1−((4sin^2 α)/9)))=4−2cosα ⇒9(1−((4sin^2 α)/9))=16−16cosα+4cos^2 α ⇒16cosα=7+4(cos^2 α+sin^2 α) ⇒cosα=((11)/(16)) ⇒α=arcos(((11)/(16))) ⇒sinβ=(2/3)(√(1−cos^2 α))=(2/3)(√(1−((121)/(256)))) =(2/3)(√((135)/(256)))=((√(135))/(24)) ⇒β=arcsin(((√(135))/(24)))

$$\Sigma{F}_{{x}} :\:\mathrm{300sin}\beta=\mathrm{200sin}\alpha\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{sin}\beta=\frac{\mathrm{2sin}\alpha}{\mathrm{3}} \\ $$$$\Sigma{F}_{\mathrm{y}} :\:\mathrm{300cos}\beta+\mathrm{200cos}\alpha=\mathrm{400} \\ $$$$\Rightarrow\mathrm{3}\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \beta}+\mathrm{2cos}\alpha=\mathrm{4} \\ $$$$\Rightarrow\mathrm{3}\sqrt{\mathrm{1}−\frac{\mathrm{4sin}^{\mathrm{2}} \alpha}{\mathrm{9}}}=\mathrm{4}−\mathrm{2cos}\alpha \\ $$$$\Rightarrow\mathrm{9}\left(\mathrm{1}−\frac{\mathrm{4sin}^{\mathrm{2}} \alpha}{\mathrm{9}}\right)=\mathrm{16}−\mathrm{16cos}\alpha+\mathrm{4cos}^{\mathrm{2}} \alpha \\ $$$$\Rightarrow\mathrm{16cos}\alpha=\mathrm{7}+\mathrm{4}\left(\mathrm{cos}^{\mathrm{2}} \alpha+\mathrm{sin}^{\mathrm{2}} \alpha\right) \\ $$$$\Rightarrow\mathrm{cos}\alpha=\frac{\mathrm{11}}{\mathrm{16}}\:\Rightarrow\alpha=\mathrm{arcos}\left(\frac{\mathrm{11}}{\mathrm{16}}\right) \\ $$$$\Rightarrow\mathrm{sin}\beta=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\mathrm{1}−\mathrm{cos}^{\mathrm{2}} \alpha}=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\mathrm{1}−\frac{\mathrm{121}}{\mathrm{256}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\sqrt{\frac{\mathrm{135}}{\mathrm{256}}}=\frac{\sqrt{\mathrm{135}}}{\mathrm{24}} \\ $$$$\Rightarrow\beta=\mathrm{arcsin}\left(\frac{\sqrt{\mathrm{135}}}{\mathrm{24}}\right) \\ $$

Answered by mr W last updated on 21/Feb/23

Commented by mr W last updated on 21/Feb/23

cos α=((2^2 +4^2 −3^2 )/(2×2×4))=((11)/(16)) ⇒α=cos^(−1) ((11)/(16))≈46.567° cos β=((3^2 +4^2 −2^2 )/(2×3×4))=(7/8) ⇒β=cos^(−1) (7/8)≈28.955°

$$\mathrm{cos}\:\alpha=\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }{\mathrm{2}×\mathrm{2}×\mathrm{4}}=\frac{\mathrm{11}}{\mathrm{16}} \\ $$$$\Rightarrow\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{11}}{\mathrm{16}}\approx\mathrm{46}.\mathrm{567}° \\ $$$$\mathrm{cos}\:\beta=\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{2}^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}×\mathrm{4}}=\frac{\mathrm{7}}{\mathrm{8}} \\ $$$$\Rightarrow\beta=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{7}}{\mathrm{8}}\approx\mathrm{28}.\mathrm{955}° \\ $$

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