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:

$c^2=6^2+17^2$

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

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

What is cosΘcosΘ if tanΘ = 6/17tanΘ = 6/17?