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  (523.74 lb)∠-48.59°

Explanation:

a) We can find the sum of the force vectors several ways, but first we need to know the direction of vector BT. The angle PBT is identical to the angle shown as α, which we can find from the lengths BD and AD.

  α = arctan(BD/AD) = arctan((3√3)/6)) ≈ 40.893°

Using the Law of Cosines, we can find the magnitude of PT from ...

  PT^2 = PB^2 +BT^2 -2·PB·BT·cos(α)

  PT^2 = 800^2 +600^2 -2·800·600·2/√7 = 1000000 -1920000/√7

  PT ≈ 523.74 . . . . lb

The direction of PT can be found from the Law of Sines.

  ∠BPT = arcsin(BT/PT·sin(α)) ≈ 1.145597sin(40.893°) ≈ 48.59°

Relative to the +x axis, the resultant force (R) is ...

  R = (523.74 lb)∠-48.59°

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b) The graphical solution is shown in the attachment. The graphing tool measures the segment lengths and angles. Those measures confirm the above result. (The pound values shown are scaled up by a factor of 100 from the segment lengths on the diagram. They are "captions" for the respective vectors.)

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