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Mathematics 22 Online
OpenStudy (anonymous):

Look at the figure. If cosec x degree is equal to 3 by 2, what is the value of y? 1 by 2 1 by 4 4 by 3 8 by 3

OpenStudy (anonymous):

@mayankdevnani this last one please :) i need a little help

OpenStudy (mayankdevnani):

if \[\huge \bf cosecx^\circ=\frac{3}{2}\] and \[\huge \bf cosec^\circ=\frac{1}{\sin^\circ}\]

OpenStudy (mayankdevnani):

so what is the value of sin? @brianjr227

OpenStudy (anonymous):

2

OpenStudy (mayankdevnani):

no

OpenStudy (anonymous):

sorry i meant 3*

OpenStudy (anonymous):

pellet

OpenStudy (anonymous):

i keep putting in the wrong stuff

OpenStudy (mayankdevnani):

\[\huge \bf sinx^\circ=\frac{2}{3}\]

OpenStudy (mayankdevnani):

okk ??

OpenStudy (anonymous):

but Y wouldn't it be 4/3 if its opposite/hypotenuse??

OpenStudy (mayankdevnani):

because its cosec.. we have to find sin.

OpenStudy (mayankdevnani):

understood? @brianjr227

OpenStudy (anonymous):

that is sin though

OpenStudy (mayankdevnani):

yeah. now clear

OpenStudy (anonymous):

oh ok! i see what you are saying, i was trying to say I found Y by using Sin function like you told me to @mayankdevnani

OpenStudy (mayankdevnani):

yeah. now clear. can we go on?

OpenStudy (anonymous):

yes lol

OpenStudy (mayankdevnani):

good.

OpenStudy (mayankdevnani):

so, \[\huge \bf sinx^\circ=\frac{2}{3}\] \[\huge \bf sinx^\circ=\frac{perpendicular=??}{hypotenuse=??}\]

OpenStudy (mayankdevnani):

can you figure out perpendicular side and hypotenuse? @brianjr227

OpenStudy (anonymous):

4/3

OpenStudy (anonymous):

?

OpenStudy (mayankdevnani):

good..

OpenStudy (mayankdevnani):

so, \[\huge \bf sinx^\circ=\frac{4y}{3}\] \[\huge \bf \frac{2}{3}=\frac{4y}{3}\] solve for y.

OpenStudy (goformit100):

@mayankdevnani Good Job.

OpenStudy (mayankdevnani):

thank you :) @goformit100

OpenStudy (anonymous):

8/3?

OpenStudy (anonymous):

@mayankdevnani

OpenStudy (anonymous):

@goformit100 8/3?

OpenStudy (goformit100):

Yes it is correct

OpenStudy (anonymous):

@goformit100 @mayankdevnani thanks :)

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