103 Trigonometry Problems

̸
1. Trigonometric Fundamentals 7
implying that
|AC| =
w
sin θ(1 + cos 2θ) .
A
B
F
E
Figure 1.7.
C
D
Second Solution: Let F be the foot of the perpendicular line segment from A to
the opposite edge. Then in the right triangle AEF , ̸ AEF = 2θ and |AF |=w.
Thus |AF |=|AE| sin 2θ, or|AE| =
sin w
2θ . In the right triangle AEC, ̸ CAE =
CAB = θ and |AE| =|AC| cos θ. Consequently,
|AC| = |AE|
cos θ = w
sin 2θ cos θ .
Putting these two approaches together, we have
|AC| =
w
sin θ(1 + cos 2θ) = w
sin 2θ cos θ ,
or sin θ(1 + cos 2θ) = sin 2θ cos θ. Interested readers can use the formulas we
developed earlier to prove this identity.
Example 1.2. In the trapezoid ABCD (Figure 1.8), AB ‖ CD, |AB| =4 and
|CD| =10. Suppose that lines AC and BD intersect at right angles, and that lines
BC and DA, when extended to point Q, form an angle of 45 ◦ . Compute [ABCD],
the area of trapezoid ABCD.