anonymous
  • anonymous
given the triangle below find sec x tan x and cos z in fraction form xz 25 yz 15
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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NotTim
  • NotTim
I don't know the answer, but you should post the triangle for other people.
anonymous
  • anonymous
|dw:1433534714159:dw|
anonymous
  • anonymous
#nottim

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More answers

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

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