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In triangle VWX shown, VX = sqrt13 and VW = 3. What is tan V?

 

 Mar 13, 2021

Best Answer 

 #1
avatar+207 
+4

Hey there!

 

First use pythagoras to gain the length of \(WX\)

 

\(WX=\sqrt{(\sqrt{13})^2-3^3}=2\)

 

Now we can take the tangent of \(V\)

 

\(tan(V)=\frac{\sqrt{(\sqrt{13})^2-3^3}}{3}=\frac{2}{3}\)

 

Hope this helped :)

 

Edit: Had \(13^2\) instead of \(\sqrt{13}^2\) oops

 Mar 13, 2021
edited by lhyla02  Mar 13, 2021
edited by lhyla02  Mar 13, 2021
edited by lhyla02  Mar 13, 2021
 #1
avatar+207 
+4
Best Answer

Hey there!

 

First use pythagoras to gain the length of \(WX\)

 

\(WX=\sqrt{(\sqrt{13})^2-3^3}=2\)

 

Now we can take the tangent of \(V\)

 

\(tan(V)=\frac{\sqrt{(\sqrt{13})^2-3^3}}{3}=\frac{2}{3}\)

 

Hope this helped :)

 

Edit: Had \(13^2\) instead of \(\sqrt{13}^2\) oops

lhyla02 Mar 13, 2021
edited by lhyla02  Mar 13, 2021
edited by lhyla02  Mar 13, 2021
edited by lhyla02  Mar 13, 2021

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