Problem:
In β³ABC, we have β C=3β 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, sinA27β=sin3A48β. Using the identity sin3A=3sinAβ4sin3A, we have
2748β=916β=sinAsin3Aβ=3β4sin2A
Solving for sinA gives sinA=11β/6 and cosA=5/6. ( cosA cannot be negative since 0<3A<180β.) Again by the Law of Sines,
sin(180ββ4A)bβ=sinA27β or b=sinA27sin4Aβ
Since
sin4A=2sin2Acos2A=4sinAcosA(cos2Aβsin2A)
we have b=27β 4β 65β(3625β11β)=35.
OR
Divide LC into Ξ±=β A and 2Ξ± as shown in the figure. By the Exterior Angle Theorem, β CDB=2Ξ±. Thus DB=CB=27 and AD=48β27=21. Since β³ADC 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,
