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Mathematics 12 Online
OpenStudy (bloomlocke367):

Use the Law of Sines. Find m

OpenStudy (bloomlocke367):

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OpenStudy (sparrow2):

m<c?

OpenStudy (bloomlocke367):

yes, the measure of angle C.

OpenStudy (bloomlocke367):

@Nnesha @perl @Preetha

OpenStudy (sparrow2):

so we need ti find sin24

OpenStudy (bloomlocke367):

but it's not a right triangle.

OpenStudy (bloomlocke367):

does that still apply?

OpenStudy (bloomlocke367):

I don't understand the Law of Sines

OpenStudy (sparrow2):

yes it still apply just write sin24

OpenStudy (bloomlocke367):

okay

OpenStudy (bloomlocke367):

I got 0.4067....

OpenStudy (bloomlocke367):

@sparrow2

OpenStudy (sparrow2):

so just enter this in the formula and then round nearest tenth

OpenStudy (bloomlocke367):

what formula?

OpenStudy (bloomlocke367):

@perl

OpenStudy (sparrow2):

law of sines

OpenStudy (bloomlocke367):

I don't know the formula.

OpenStudy (sparrow2):

162/sinc=77/sin24

OpenStudy (sparrow2):

then find sinc and look at the table to the relevant angle

OpenStudy (bloomlocke367):

what table? and I'm very confused right now..

OpenStudy (sparrow2):

values of trig function at different angle

OpenStudy (perl):

Does it make sense up to here $$ \Large \frac{\sin 24^o}{77}= \frac{\sin C}{162} $$

OpenStudy (bloomlocke367):

why is it like that?

OpenStudy (bloomlocke367):

ohhhh, yes that makes sense.

OpenStudy (perl):

so we are using just two parts of the three part ratio

OpenStudy (perl):

you can do that since they are equal

OpenStudy (bloomlocke367):

okay.

OpenStudy (perl):

we don't know `sin B` and `b` anyway at this point , yet

OpenStudy (bloomlocke367):

and we have to solve for C?

OpenStudy (perl):

$$ \Large{ \frac{\sin 24^o}{77}= \frac{\sin C}{162} \\~\\ \iff \\~\\ \sin 24^o \times 162 = 77 \times \sin C \\~\\ \iff \\~\\ \frac{\sin 24^o \times 162}{77} = \sin C }$$

OpenStudy (bloomlocke367):

okay. and how do we isolate C?

OpenStudy (perl):

$$ \Large{ \frac{\sin 24^o}{77}= \frac{\sin C}{162} \\~\\ \iff \\~\\ \sin 24^o \times 162 = 77 \times \sin C \\~\\ \iff \\~\\ \frac{\sin 24^o \times 162}{77} = \sin C \\~\\ \iff \\~\\ \sin^{-1} \left( \frac{\sin 24^o \times 162}{77} \right)= C }$$

OpenStudy (bloomlocke367):

that's what I thought, okay.

OpenStudy (perl):

make sure you are in degree mode if you are using a TI 83/84/85 calculator

OpenStudy (bloomlocke367):

I got 58.8

OpenStudy (bloomlocke367):

is that right?

OpenStudy (perl):

yes :)

OpenStudy (perl):

|dw:1432757526103:dw|

OpenStudy (bloomlocke367):

Oh, wow. Thank you so much! I might need your help later, but I don't know. \(\Huge\rlap{\tt\color{teal}{Thank~you!}}{\;\tt\color{lime}{Thank~you!}}\)

OpenStudy (perl):

Now try to 'solve' the triangle (find all side lengths and angles). You can find angle B using the fact that three angles of any triangle add up to 180. Then use law of sines one more time to find side b, or alternatively you can use law of cosines

OpenStudy (perl):

Your welcome :)

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