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1 solution
Solution 1:
By Letting:
x = angle A
2x = Angle B
4x = Angle C
By Law of sines,
a/(sin A) = b/(sin B) = c/(sin C)
= a/(sin x) = b/(sin 2x) = c/(sin 4x)
So,,
b/a = (sin 2x)/(sin x) = {[2(sin x)(cos x)]}/(sin x) = 2(cos x)
c/b = (sin 4x)/(sin 2x) = {[2(sin 2x)(cos 2x)]}/(sin 2x) = 2(cos 2x) = 2[(cos^2)(x) - (sin^2)(x)]
Therefore,
(b/a)^2 - (c/b)
= {[2(cos x)]^2} - {2[(cos^2)(x) - (sin^2)(x)]}
= [4(cos^2)(x)] - {2[(cos^2)(x) - (sin^2)(x)]}
= [4(cos^2)(x)] - [2(cos^2)(x) - 2(sin^2)(x)]
= 2(cos^2)(x) + 2(sin^2)(x)
= 2
Final Answer: 2
Solution 2:
a=2RsinA,b=2RsinB,c=2RsinC
sin^2(B)/sin^2(A)-sinC/sinB
substituting the values and using the identity sin 2X=2sinX cos X in both fractions we get
4cos^2(A)-2cos2A
=4cos^2(A)-2(2cos^2(A)-1)=2
; where in, R is the Circum-radius
Solution 3:
By law of sines,
b/a = sin(2A)/sin(A) = 2*cos(A) and
c/b = sin(4A)/sin(2A) = 2*cos(2A).
Thus,
(b/a)^2 - c/b
= 4 * cos^2(A) - 2 * cos(2A)
= 4 * cos^2(A)- [4*cos^(A)-2]
= 2..