[MP PET 1989]
A) 1/2
B) 1/4
C) 1/6
D) 1/8
Correct Answer: D
$ \cos 20^{o}\cos 40^{o}\cos 80^{o}=\frac{\sin 2^{3}20^{o}}{2^{3}\sin 20^{o}} $ $ =\frac{\sin 160^{o}}{8\sin 20^{o}}=\frac{1}{8} $ .
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