Question-116367

Question Number 116367 by soumyasaha last updated on 03/Oct/20

Answered by mr W last updated on 03/Oct/20

(T_1 /(sin 58°))=(T_2 /(sin 40°))=((100)/(sin (50°+32°))) ⇒T_1 =((sin 58° ×100)/(sin 82°))=85.64 lb ⇒T_2 =((sin 40° ×100)/(sin 82°))=64.91 lb

$$\frac{{T}_{\mathrm{1}} }{\mathrm{sin}\:\mathrm{58}°}=\frac{{T}_{\mathrm{2}} }{\mathrm{sin}\:\mathrm{40}°}=\frac{\mathrm{100}}{\mathrm{sin}\:\left(\mathrm{50}°+\mathrm{32}°\right)} \\ $$$$\Rightarrow{T}_{\mathrm{1}} =\frac{\mathrm{sin}\:\mathrm{58}°\:×\mathrm{100}}{\mathrm{sin}\:\mathrm{82}°}=\mathrm{85}.\mathrm{64}\:{lb} \\ $$$$\Rightarrow{T}_{\mathrm{2}} =\frac{\mathrm{sin}\:\mathrm{40}°\:×\mathrm{100}}{\mathrm{sin}\:\mathrm{82}°}=\mathrm{64}.\mathrm{91}\:{lb} \\ $$

Commented by mr W last updated on 03/Oct/20

Answered by Dwaipayan Shikari last updated on 04/Oct/20

T_1 sin50°+T_2 sin32°=100 T_1 cos50°=T_2 cos32° (T_1 /T_2 )=((cos32°)/(cos50°)) (T_1 /T_2 )sin50°+sin32°=((100)/T_2 ) cos32°tan50°+sin32°=((100)/T_2 ) T_2 =64.912 lb T_1 =T_2 .((cos32°)/(cos50°))=85.63 lb

$$\mathrm{T}_{\mathrm{1}} \mathrm{sin50}°+\mathrm{T}_{\mathrm{2}} \mathrm{sin32}°=\mathrm{100} \\ $$$$\mathrm{T}_{\mathrm{1}} \mathrm{cos50}°=\mathrm{T}_{\mathrm{2}} \mathrm{cos32}° \\ $$$$\frac{\mathrm{T}_{\mathrm{1}} }{\mathrm{T}_{\mathrm{2}} }=\frac{\mathrm{cos32}°}{\mathrm{cos50}°} \\ $$$$\frac{\mathrm{T}_{\mathrm{1}} }{\mathrm{T}_{\mathrm{2}} }\mathrm{sin50}°+\mathrm{sin32}°=\frac{\mathrm{100}}{\mathrm{T}_{\mathrm{2}} } \\ $$$$\mathrm{cos32}°\mathrm{tan50}°+\mathrm{sin32}°=\frac{\mathrm{100}}{\mathrm{T}_{\mathrm{2}} } \\ $$$$\mathrm{T}_{\mathrm{2}} =\mathrm{64}.\mathrm{912}\:\mathrm{lb} \\ $$$$\mathrm{T}_{\mathrm{1}} =\mathrm{T}_{\mathrm{2}} .\frac{\mathrm{cos32}°}{\mathrm{cos50}°}=\mathrm{85}.\mathrm{63}\:\mathrm{lb} \\ $$

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