Use the Law of Sines to solve for the measure of angle C. A=150, a=6, c=4
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would it be 4sin150=6sinC
yeap
i got 2=6sinC and don't know what to do
\(\bf \cfrac{sin(150^o)}{6}=\cfrac{sin(C)}{4}\implies \cfrac{4sin(150^o)}{6}=sin(C)\)
1/3?
yes,\(\bf \times sin(150^o)\)
1/6?
that makes no sense to me.. the other angles are whole numbers.. how is it possible C is only 1/6?
You have to take the inverse of sin to get the angle.
sorry the site is getting a bit laggy
as pointed out by coolsday , that's just the SINE function value, not the angle itself
so what do i do next?
\(\bf \cfrac{sin(150^o)}{6}=\cfrac{sin(C)}{4}\implies \cfrac{4sin(150^o)}{6}=sin(C)\implies \square =sin(C)\\ \quad \\ sin^{-1}(\square )= sin^{-1}[sin(C)]\implies sin^{-1}(\square )= \measuredangle C\)
i'm sorry... i'm still not understanding
hmmm are you using Internet Explorer by any chance?
no, google chrome
hmm ... so... .ahemm.... \(\bf \cfrac{sin(150^o)}{6}=\cfrac{sin(C)}{4}\implies \cfrac{4sin(150^o)}{6}=sin(C)\implies \square =sin(C)\\ \quad \\ {\color{red}{ sin^{-1}}}(\square )= {\color{red}{ sin^{-1}}}[sin(C)]\implies {\color{red}{ sin^{-1}}}(\square )= \measuredangle C\)
so i have to take the inverse of 1/6?
yes, the inverse of whatever the left-side yielded
well the sin inverse of 1/6
yes
i forgot my calculator at school and cant find an online calculator with inverse sin
www.desmos.com
still not seeing inverse sin
click on [Functions ] button
i did and all i see is sin cos tan csc sec cot.
hmm I guess it maybe in radians... .one sec -> https://www.google.com/search?client=opera&rls=en&q=asin(1/6+)+in+degrees&sourceid=opera&ie=utf-8&oe=utf-8&channel=suggest
so would that be the answer?
www.speedcrunch.org <--- in case you need a calculator
didnt mean to resend that
\(\bf {\color{red}{ sin^{-1}}}(\square )= \measuredangle C\implies 9.59^o\approx \measuredangle C\)
9.59 is angle c?
yeap
thank you so much!!
yw
I think you should recheck your calculations again. The sine function value is not 1/6.
i really don't know how do this problem at all
|dw:1389821643934:dw| cross multiply |dw:1389821680508:dw| The way you are solving for the angle is correct, it is just that you made a mistake in your calculations in step 1
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