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Mathematics 14 Online
OpenStudy (sup_open~study):

can some one help solve the answer please

OpenStudy (sup_open~study):

OpenStudy (sup_open~study):

my assignment restarted and i had to start all over again and i forgot almost all of it :(

OpenStudy (sup_open~study):

sin

OpenStudy (diyadiya):

What is sinM = ?

OpenStudy (sup_open~study):

no wait it's cos

OpenStudy (diyadiya):

So what is Cos(M)= ?

OpenStudy (sup_open~study):

k one sec

OpenStudy (sup_open~study):

:)

OpenStudy (diyadiya):

OH WAIT!!!!!!!!

OpenStudy (diyadiya):

Question is to find Cosecant of angle M

OpenStudy (sup_open~study):

i got 0.999999 and a lot more stuff

OpenStudy (diyadiya):

@Sup_open~study you don't have to convert it into decimals

OpenStudy (sup_open~study):

one sec i'll be back

OpenStudy (sup_open~study):

o ok i messed some were i got to reset my calculator

OpenStudy (sup_open~study):

it has to do on the d i think is that right ???

OpenStudy (diyadiya):

Nope d is not the answer

OpenStudy (diyadiya):

\[Cosecant \theta= \frac{Hypotenuse}{Opposite \ side}\]

OpenStudy (diyadiya):

use the pythagorus theorem to find the oppsite side PN

OpenStudy (sup_open~study):

o crap i had it the other way lol i got this now xD

OpenStudy (diyadiya):

PM^2 = PN^2 + MN^2 PN^2= PM^2 -MN^2 =17^2-8^2 SO YOU'LL GET PM =15

OpenStudy (diyadiya):

Opposite side =15 Hypotenuse = 17

OpenStudy (diyadiya):

Now you can do for cosecant

OpenStudy (sup_open~study):

i hate this part xD

OpenStudy (diyadiya):

cosecant = hyp/opp = ?

OpenStudy (diyadiya):

17/15

OpenStudy (diyadiya):

Sorry for the confusion :S

OpenStudy (sup_open~study):

yeah but ins't it like -cos

OpenStudy (diyadiya):

Cos & Cosecant is different

OpenStudy (diyadiya):

cosecant = 1/sin

OpenStudy (diyadiya):

secant = 1/cos

OpenStudy (sup_open~study):

what do you mean by 1/sin ???

OpenStudy (sup_open~study):

thank you for the help :)

OpenStudy (sup_open~study):

can you help ??? i'm stuck

OpenStudy (anonymous):

|dw:1337371580382:dw| i think this would be ur answer if u r looking for csc

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