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

- anonymous

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

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- schrodinger

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- NotTim

I don't know the answer, but you should post the triangle for other people.

- anonymous

|dw:1433534714159:dw|

- anonymous

#nottim

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## More answers

- Nnesha

@ <<---- to tag some
@name

- Nnesha

@ <<---- to tag some
@ name

- Nnesha

|dw:1433535491052:dw|
it's right triangle right ????

- Nnesha

\[\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }\]

- 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

ok

- anonymous

@Nnesha

- Nnesha

^ right
so what did you got for 3rd side ?

- anonymous

im really behind and i just need answers
i Know it sound rude but im jugling 10 assignment

- anonymous

at the same time

- Nnesha

welcome to openstudy!
that's not gonna happen you have to participate :-)

- Nnesha

alright substitute a and c value solve for b :-)

- Nnesha

\[\huge\rm 15^2 + b^2 = 25^2\]
b = ?

- anonymous

90

- anonymous

@Nnesha

- Nnesha

nope how did you get 90 ?

- Nnesha

take square of 25 and 15

- anonymous

625 and 225

- Nnesha

\[\huge\rm a^2+ 225 = 625\]
solve for a
move 225 to the right side

- anonymous

then subtract 225

- Nnesha

yes right

- Nnesha

then take square root

- anonymous

20

- Nnesha

ys right

- 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

can we do another'

- Nnesha

did you get that one ?
cos = ?
tan = ?

- Nnesha

remember you need
tan x
and cos z

- anonymous

i got it

- Nnesha

alright

- anonymous

|dw:1433539551579:dw|

- anonymous

@Nnesha

- Nnesha

sin^ theta =what ?
remember the identity ?

- Nnesha

sin^2 theta =what ?
remember the identity ?

- anonymous

its a different question

- Nnesha

yes ik :-)

- anonymous

|dw:1433539731456:dw|

- anonymous

i have to simplify

- 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

no

- anonymous

i realy need the answer i only have a couple hours to turn in everything

- 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

ok i understand

- anonymous

this is a different one

- anonymous

|dw:1433540518518:dw|

- anonymous

@Nnesha

- anonymous

ill be serious

- Nnesha

alright what's the statement ?

- anonymous

angle j

- anonymous

?

- Nnesha

is it right triangle ?

- anonymous

no

- anonymous

is it 35degrees? @Nnesha

- Nnesha

did you apply law of cosines or law of sines ?

- anonymous

cosine

- anonymous

was it right?

- Nnesha

gimme a sec i'll check

- Nnesha

law of cosines when they gave you one angle
check your question again

- anonymous

then its 7

- Nnesha

i din't say 35 is wrong ....
for law of cosine
you should have all 3 sides or at least one angle

- anonymous

thats the angle tho

- Nnesha

what's the question
seems like you mssing something
look at the original question

- Nnesha

are*

- anonymous

what is the measure of angle j in the triangle

- Nnesha

take a screenshot

- 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

|dw:1433541529143:dw|

- Nnesha

yep now we can solve it
thanks!

- anonymous

sorry

- Nnesha

nope 35 isn't right
you have to use law of sines

- Nnesha

\[\huge\rm \frac{ \sin J}{ j } = \frac{ \sin I }{ i }\]
plug in numbers solve for J

- Nnesha

j is opposite side of angle J|dw:1433541845296:dw|

- anonymous

\[(\sin 10/10)\]

- Nnesha

sin is not 10
j side = 10

- Nnesha

\[\huge\rm \frac{ \sin J}{ j } = \frac{ \sin I }{ i }\]
angle I = ?
side i = ?
j side = 10

- anonymous

i = 17

- Nnesha

yes right so plug in

- Nnesha

and angle I = ?

- anonymous

102

- Nnesha

yes right

- Nnesha

\[\huge\rm \frac{ \sin J}{ 10 } = \frac{ \sin (102) }{ 17 }\]
solve for angle J

- anonymous

83

- Nnesha

i didn't get 83

- anonymous

oh sorry 97

- Nnesha

i didn't get 97

- anonymous

what did you get?

- Nnesha

how did you solve ?
show your work please :(

- anonymous

im new to this it will take a long tim

- anonymous

7

- Nnesha

it's okay

- Nnesha

how did you get 7 ? :o

- Nnesha

did you use calculator ?

- anonymous

yes

- Nnesha

what did u put into the calculator ?
parentheses are really important

- anonymous

sin/ 10=sin(102)/17

- Nnesha

sin J**

- Nnesha

http://www.wolframalpha.com/ use this one

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