Complete question
The diagram for this question is shown on the first uploaded image
Answer:
[tex]T_a = 44.8 \ N [/tex] , [tex] F = 86.03 \ N [/tex]
Explanation:
From the question we are told that
The magnitude of [tex]T_b = 80 N[/tex]
From the diagram we can see that
[tex]Ta * sin (38) + F * sin (29) = T_b * cos (30) \ \cdots (1)[/tex]
[tex]Ta * sin (38) + F * sin (29) = 80 * cos (30) \ \cdots (1)[/tex]
[tex]0.616 Ta + 0.485F = 69.3 \ \cdots (1)[/tex]
=> [tex]T_a = \frac{69.3 - 0.485F}{0.616}[/tex]
Also
[tex]T_a * cos(38) + T_b * sin (30)= F * cos (29) \ \cdots (2)[/tex]
=> [tex]T_a * cos(38) + 80* sin (30)= F * cos (29) \ \cdots (2)[/tex]
=> [tex]0.788 T_a + 40 = 0.875 F \ \cdots (2)[/tex]
=> [tex]0.788 [\frac{69.3 - 0.485F}{0.616}]+ 40 = 0.875 F[/tex]
=> [tex] 88.65 - 0.6204 F + 40 = 0.875 F[/tex]
=> [tex] 88.65 + 40 = 0.875 F+0.6204 F [/tex]
=> [tex] 128.65 = 1.4954 F [/tex]
=> [tex] F = 86.03 \ N [/tex]
substituting this obtained value for F in the above equation
[tex]T_a = \frac{69.3 - 0.485(86.03)}{0.616}[/tex]
[tex]T_a = 44.8 \ N [/tex]
