given the triangle below find sec x tan x and cos z in fraction form xz 25 yz 15

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given the triangle below find sec x tan x and cos z in fraction form xz 25 yz 15

Mathematics
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I don't know the answer, but you should post the triangle for other people.
|dw:1433534714159:dw|
#nottim

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|dw:1433535491052:dw| it's right triangle right ????
\[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]
yes it's a right triangle find 3rd side of right triangle Pythagorean theorem \[\huge\rm a^2 + b^2 = c^2\] c is the longest side of right triangle which is hypotenuse solve for 3rd side
ok
^ right so what did you got for 3rd side ?
im really behind and i just need answers i Know it sound rude but im jugling 10 assignment
at the same time
welcome to openstudy! that's not gonna happen you have to participate :-)
alright substitute a and c value solve for b :-)
\[\huge\rm 15^2 + b^2 = 25^2\] b = ?
90
nope how did you get 90 ?
take square of 25 and 15
625 and 225
\[\huge\rm a^2+ 225 = 625\] solve for a move 225 to the right side
then subtract 225
yes right
then take square root
20
ys right
|dw:1433537872991:dw| now use this information \[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\] cos = adjacent over hyp tan = opposite over adjacent remember sec = 1/cos
can we do another'
did you get that one ? cos = ? tan = ?
remember you need tan x and cos z
i got it
alright
|dw:1433539551579:dw|
sin^ theta =what ? remember the identity ?
sin^2 theta =what ? remember the identity ?
its a different question
yes ik :-)
|dw:1433539731456:dw|
i have to simplify
ik i'm asking do you know the sin cos identities most important one is \[\huge\rm sin^2 \theta + \cos^2 =1\] solve it for sin^2
no
i realy need the answer i only have a couple hours to turn in everything
sin^2 + cos ^2 = 1 solve for sin^2 please first of all i don't like giving direct answers (enabling ) 2nd i'm not allowed to give you direct answer http://openstudy.com/code-of-conduct
ok i understand
this is a different one
|dw:1433540518518:dw|
ill be serious
alright what's the statement ?
angle j
?
is it right triangle ?
no
is it 35degrees? @Nnesha
did you apply law of cosines or law of sines ?
cosine
was it right?
gimme a sec i'll check
law of cosines when they gave you one angle check your question again
then its 7
i din't say 35 is wrong .... for law of cosine you should have all 3 sides or at least one angle
thats the angle tho
what's the question seems like you mssing something look at the original question
are*
what is the measure of angle j in the triangle
take a screenshot
i'm saying they should gave u all 3 sides or 2 sides one angle that's not info to solve the question
|dw:1433541529143:dw|
yep now we can solve it thanks!
sorry
nope 35 isn't right you have to use law of sines
\[\huge\rm \frac{ \sin J}{ j } = \frac{ \sin I }{ i }\] plug in numbers solve for J
j is opposite side of angle J|dw:1433541845296:dw|
\[(\sin 10/10)\]
sin is not 10 j side = 10
\[\huge\rm \frac{ \sin J}{ j } = \frac{ \sin I }{ i }\] angle I = ? side i = ? j side = 10
i = 17
yes right so plug in
and angle I = ?
102
yes right
\[\huge\rm \frac{ \sin J}{ 10 } = \frac{ \sin (102) }{ 17 }\] solve for angle J
83
i didn't get 83
oh sorry 97
i didn't get 97
what did you get?
how did you solve ? show your work please :(
im new to this it will take a long tim
7
it's okay
how did you get 7 ? :o
did you use calculator ?
yes
what did u put into the calculator ? parentheses are really important
sin/ 10=sin(102)/17
sin J**
http://www.wolframalpha.com/ use this one

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