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rebeccaskell94
Given the triangle below, what is cot∡B? http://learn.flvs.net/webdav/assessment_images/educator_geometry_v14/pool_Geom_3641_0403_03/image0084e2ec0b0.gif √33/ 4 7•√33/ 33 √33 /7 4•√33/33 I think the Answer is C?
Use Pythagoras theorem
I think that is a) \(\sqrt{33}/4\)
urg I couldn't figure out if it was a or c because I wasn't sure if 4 or 7 should be the denominator
cot B = \(\frac{1}{tanB}\) right ?
@rebeccaskell94 First step is to find AB we know that using Pythagoras \[BC^2=AB^2+AC^2\] Find AB from here \[\cot =\frac{adjacent}{perpendicular}=\frac1 {\tan}\]
tan B = opposite/adjacent opposite = 4 adjacent = ??? We will find adj. by pythagoras theorem
\[base^2+perpendicular^2=hypotenuse^2\] Adjacent = \(\sqrt{33}\) hence we have tan B = \(\frac{\sqrt{33}}{4}\)
did it help @rebeccaskell94 ?
Yes! Very much so. I'm writing it down now so I remember the basic idea :) Thank you so much!
Nice to hear that @rebeccaskell94 :)
See http://www.khanacademy.org/math/trigonometry/v/basic-trigonometry for a video on sin,cos, tan defs using SOHCAHTOA a the co functions are harder to remember!
SOH CAH TOA much better sine = opposite / hypotenuse cosine = adj. / hypotenuse tan = opposite / adj. right ?