Wednesday

July 30, 2014

July 30, 2014

Posted by **Mark** on Tuesday, March 23, 2010 at 4:47pm.

sec Q = 5 tan = 2sqrt6

a) cos Q b) cotQ

c) cot(90 degrees - Q) d) sin Q

Please explain. I do not understand how to do this. Thank you!!

- Math -
**Reiny**, Tuesday, March 23, 2010 at 5:22pmsec Q = 5 tanQ

1/cosQ = 5sinQ/cosQ

so sinQ = 1/5

I then made a right angles triangle with hypotenuse 5, opposite 1 and let x be the adjacent.

By Pythagoras

x^2 + 1^2 = 5^2

x^2 = 24

x = √24 = 2√6

WOAH!

but you said secQ = 2√6

if, as you started, secQ = 5tanQ, then my calculations stands as correct and

secQ would be 5/(2√6) and not 2√6

I will wait till you correct your opening statement of

sec Q = 5 tan = 2sqrt6 , which leads to a contradiction.

- Math -
**Aaron**, Wednesday, July 20, 2011 at 1:54amWhat this person means is :

cosQ = 5 and tanQ = 2sqrt6

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