Question

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?

Answer 1

Use Pythagoras theorem

Answer 2

I think that is a) \(\sqrt{33}/4\)

Answer 3

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

Answer 4

cot B = \(\frac{1}{tanB}\) right ?

Answer 5

@rebeccaskell94 First step is to find AB
we know that using Pythagoras
\[BC^2=AB^2+AC^2\]
Find AB from here
\[\cot =\frac{adjacent}{perpendicular}=\frac1 {\tan}\]

Answer 6

tan B = opposite/adjacent
opposite = 4
adjacent = ???

We will find adj. by pythagoras theorem

Answer 7

\[base^2+perpendicular^2=hypotenuse^2\]

Adjacent = \(\sqrt{33}\)

hence we have tan B = \(\frac{\sqrt{33}}{4}\)

Answer 8

did it help @rebeccaskell94 ?

Answer 9

Yes! Very much so. I'm writing it down now so I remember the basic idea :) Thank you so much!

Answer 10

Nice to hear that @rebeccaskell94 :)

Answer 11

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!

Answer 12

SOH CAH TOA much better

sine = opposite / hypotenuse

cosine = adj. / hypotenuse

tan = opposite / adj.

right ?