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

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

$$\frac{T_1}{\sin 58°}=\frac{T_2}{\sin 40°}=\frac{100}{\sin (50°+32°)}$$$$\Rightarrow T_1=\frac{\sin 58°\times 100}{\sin 82°}=85.64lb$$$$\Rightarrow T_2=\frac{\sin 40°\times 100}{\sin 82°}=64.91lb$$

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

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

$${\mathrm{T}}_1\sin 50°+{\mathrm{T}}_2\sin 32°=100$$$${\mathrm{T}}_1\cos 50°={\mathrm{T}}_2\cos 32°$$$$\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}=\frac{\cos 32°}{\cos 50°}$$$$\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}\sin 50°+\sin 32°=\frac{100}{{\mathrm{T}}_2}$$$$\cos 32°\tan 50°+\sin 32°=\frac{100}{{\mathrm{T}}_2}$$$${\mathrm{T}}_2=64.912\mathrm{lb}$$$${\mathrm{T}}_1={\mathrm{T}}_2.\frac{\cos 32°}{\cos 50°}=85.63\mathrm{lb}$$

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