What is #cosΘ# if #tanΘ = 6/17#?

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Answer 1 · 2018-04-01T15:07:53

#=> (17sqrt(13) ) / 65 #

We know #tantheta = "opposite"/"adjacent" #

Hence #opposite" = 6 # and #"adjacent" = 17 #

Hence using Pythagorus rule: #a^2 + b^2 = c^2 #

#=> 6^2 + 17^2 = c^2 => c = 5sqrt(13) #

#=> "hypotenuse" = 5sqrt(13) #

We know #costheta = "adjacent" / "hypotenuse" #

#=> cos theta = 17/(5sqrt(13 ) ) #

#=> (17sqrt(13) ) / 65 #

Answer 2 · 2018-04-01T15:21:51

#color(blue)((17sqrt(13))/65)#

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From diagram, we have:

#tan(theta)="opposite"/"adjacent"=6/17#

#cos(theta)="adjacent"/"hypotenuse"=17/c#

We need to find #bbc#:

Using Pythagoras' Theorem:

#c^2=6^2+17^2#

#c=sqrt((6)^2+(17)^2)=sqrt(325)=5sqrt(13)#

#cos(theta)="adjacent"/"hypotenuse"=17/(5sqrt(13))=color(blue)((17sqrt(13))/65)#