ABC and BCD are two attached triangles, attached along line BC. AB = 5.8cm, AC=5.2cm, BD=4.3cm. Angle BDC = 30 degrees, and angle DCB is a right angle. Calculate angle CBA.

This question teaches the importance of drawing diagrams to figure out what a question is asking, as well as deeply testing how well the student understands trigonometry and equation rearrangement, combining all these techniques into one question. As BC is shared by both triangles ABC and BCD, the first step should be to calculate length BC using sin(30) = BC/4.3 in the right angle triangle, giving BC=4.3sin(30)=2.15cm. Next, as we have 3 sides and no angles to triangle ABC, we must rearrange the cosine rule equation to form cos(A) = (b^2 + c^2 -a^2 / 2bc). Then, place the correct values into the equation and calculate cos(A), in this case cos(A)=((5.8^2)+(2.15^2)-(5.2^2))/(25.82.15). When you apple inverse cosine on the result from this, the angle calculated is 63.3 degrees (3s.f.)