Problem:
In △ABC, we have ∠C=3∠A,a=27 and c=48. What is b?
Answer Choices:
A. 33
B. 35
C. 37
D. 39
E. not uniquely determined
Solution:
By the Law of Sines, sinA27=sin3A48. Using the identity sin3A=3sinA−4sin3A, we have
2748=916=sinAsin3A=3−4sin2A
Solving for sinA gives sinA=11/6 and cosA=5/6. ( cosA cannot be negative since 0<3A<180∘.) Again by the Law of Sines,
sin(180∘−4A)b=sinA27 or b=sinA27sin4A
Since
sin4A=2sin2Acos2A=4sinAcosA(cos2A−sin2A)
we have b=27⋅4⋅65(3625−11)=35.
OR
Divide LC into α=∠A and 2α as shown in the figure. By the Exterior Angle Theorem, ∠CDB=2α. Thus DB=CB=27 and AD=48−27=21. Since △ADC is isosceles also, CD=21. Now apply Stewart's Theorem [see H.S.M. Coxeter & S.L. Greitzer, "Geometry Revisited," New Mathematical Library, Vol. 19], which says that for any point D on AB,
