## 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? one year ago one year ago

1. goformit100

Use Pythagoras theorem

2. mathslover

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

4. mathslover

cot B = $$\frac{1}{tanB}$$ right ?

5. ash2326

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

6. mathslover

tan B = opposite/adjacent opposite = 4 adjacent = ??? We will find adj. by pythagoras theorem

7. mathslover

$base^2+perpendicular^2=hypotenuse^2$ Adjacent = $$\sqrt{33}$$ hence we have tan B = $$\frac{\sqrt{33}}{4}$$

8. mathslover

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

10. mathslover

Nice to hear that @rebeccaskell94 :)

11. phi

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!

12. mathslover

SOH CAH TOA much better sine = opposite / hypotenuse cosine = adj. / hypotenuse tan = opposite / adj. right ?