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Mathematics 17 Online
OpenStudy (anonymous):

Trigonometry: check my work? and I need help to find sec theta?

OpenStudy (anonymous):

jimthompson5910 (jim_thompson5910):

sec(theta) = 1/cos(theta) since cos(theta) = adj/hyp, this means sec(theta) = hyp/adj you take the reciprocal of cos(theta) to find sec(theta)

jimthompson5910 (jim_thompson5910):

also, your answer of sin(theta) = -3/4 is incorrect your triangle should be labeled like this |dw:1400973476401:dw|

OpenStudy (anonymous):

would sinθ be 4/H 5^2+y^2=4^2 25+y^2=16 y= -9 -4/3 <-

jimthompson5910 (jim_thompson5910):

5^2+y^2=4^2 is the incorrect equation

jimthompson5910 (jim_thompson5910):

it should be 5^2+4^2 = y^2 y is the hypotenuse

OpenStudy (anonymous):

-4/41 ?

jimthompson5910 (jim_thompson5910):

not quite

OpenStudy (anonymous):

What do I do?

jimthompson5910 (jim_thompson5910):

what is 5^2+4^2 equal to?

OpenStudy (anonymous):

41

OpenStudy (anonymous):

41

jimthompson5910 (jim_thompson5910):

mathman123 please don't give out the answer

OpenStudy (anonymous):

sry

jimthompson5910 (jim_thompson5910):

5^2+4^2 = y^2 y^2 = 41 y = ???

OpenStudy (anonymous):

20.5?

jimthompson5910 (jim_thompson5910):

y^2 means y squared not y times 2

OpenStudy (anonymous):

Im confused do i find √41?

jimthompson5910 (jim_thompson5910):

it's just \[\Large \sqrt{41}\]

jimthompson5910 (jim_thompson5910):

|dw:1400975463702:dw|

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