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Step-by-step explanation:

In the parallelogram shown, angle B is the supplement of angle DAB:

  ∠B = 180° -77°37' = 102°23'

Angle ACB is the difference of angles 77°37' and 27°8', so is 50°29'.

Now, we know the angles and one side of triangle ABC. We can use the law of sines to solve for the other two sides.

  BC/sin(A) = AB/sin(C)

  AD = BC = AB·sin(A)/sin(C) = (169 lb)sin(27°8')/sin(50°29') ≈ 99.91 lb

 AC = AB·sin(B)/sin(C) = (169 lb)sin(102°23')/sin(50°29') ≈ 213.97 lb

step-by-step explanation: