Problem:
In rectangle ABCD,AB=20 and BC=10. Let E be a point on CD such that ∠CBE=15∘. What is AE?
Answer Choices:
A. 3203
B. 103
C. 18
D. 113
E. 20
Solution:
Let E′ be the point on CD such that AE′=AB=2AD. Then △ADE′ is a 30−60−90∘ triangle, so ∠DAE′=60∘. Hence ∠BAE′=30∘. Also, AE′=AB implies that ∠E′BA=∠BE′A=75∘, and then ∠CBE′=15∘. Thus it follows that E′ and E are the same point. Therefore, AE=AE′=AB=(E)20.
