Actually, there are 3 variables with only two equations given. So there would be no unique solutions normally. But since the sin(theta) and cos(theta) are trigonometric, one can be expressed in terms of the other so in the end, there are effectively only two variables in two equatons.

So eliminate W first.

For simplicity in my typing, let A = theta

From (1),

W = 40/sinA

Substitute that into (2),

69.28 -40/sinA -(40/sinA)cosA = 0

69.28 -40cscA -40cotA = 0

Divide both sides by 40,

1.732 -cscA -cotA = 0

1.732 -cotA = cscA

Squatre both sides,

3 -3.464cotA +cot^2(A) = csc^2(A) -----(i)

We know sin^2(A) +cos^2(A) = 1

Divide both sides by sin^2(A),

1 +cot^2(A) = csc^2(A)

Substitute that into (i),

3 -3.464cotA +cot^2(A) = 1 +cot^2(A)

2 -3.464cotA = 0

1 -1.732cotA = 0

[sqrt(3)]cotA = 1

cotA = 1 / sqrt(3)

tanA = sqrt(3)

A = 60 degrees ----------answer.

And, W = 40/sin(60deg) = 46.188 --------answer.