How do you find the exact value for #(5tan60°) / cos30°#?

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Answer 1 · 2015-06-08T06:04:32

You can do this without a calculator and with few manipulations.

#(5tan60^o)/(cos30^o) = (5sin60^o)/[(cos30^o)(cos60^o)]#

but

#sin(pi/2 - x) = cosx#

Therefore:

#sin(90^o - overbrace(30^o)^x) = sin60^o = cos30^o#

#=> (5cancel(sin60^o))/[cancel((cos30^o))(cos60^o)]#

#= (5cancel(sqrt3/2))/(cancel(sqrt3/2)*1/2)#

#= 10#