The simple truss structure is shown below. The forces acting in the members AB and BC respectively are
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To quickly solve simple right-triangle trusses, use the load to solve the vertical component first. Since the downward load \( W \) must be balanced by the vertical component of the angled strut, \( F_{BC}\sin(30^\circ) = W \implies F_{BC} = 2W \) instantly!
Step 1: Picture the three forces at joint B as a closed force triangle.
At joint B, the vertical load \( W \), the horizontal member force \( F_{AB} \), and the inclined member force \( F_{BC} \) must balance each other, so drawing them tip to tail forms a closed triangle. Since AB is horizontal and \( W \) is vertical, the angle between \( F_{AB} \) and \( W \) in this triangle is \( 90^\circ \), and \( F_{BC} \) makes a \( 30^\circ \) angle with \( F_{AB} \), the same \( 30^\circ \) that member BC makes with the horizontal.
Step 2: Read the triangle as a right triangle.
This makes the force triangle a right triangle, with \( F_{BC} \) as the hypotenuse (it is the only member force with both a horizontal and a vertical component), \( W \) as the side opposite the \( 30^\circ \) angle, and \( F_{AB} \) as the side adjacent to it.
Step 3: Use basic right-triangle trig to solve for both forces.
\[
\sin(30^\circ) = \frac{W}{F_{BC}} \quad \Rightarrow \quad F_{BC} = \frac{W}{\sin(30^\circ)} = \frac{W}{0.5} = 2W
\]
\[
\tan(30^\circ) = \frac{W}{F_{AB}} \quad \Rightarrow \quad F_{AB} = \frac{W}{\tan(30^\circ)} = \sqrt{3}\,W \approx 1.732W
\]
Member BC carries this load by pushing back into the joint to hold it up, so it is in compression, while member AB is stretched taut resisting the horizontal pull, so it is in tension.
So \( F_{AB} = \boxed{1.732W \text{ (Tensile)}} \) and \( F_{BC} = \boxed{2W \text{ (Compression)}} \), matching option (D).
