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I seriously just did a problem like this but I always forgot on how to use it

HI!!

looks like a law of cosines problem?

sine law again.

hmm i think you need cosines for this
you do not know any angle opposite one of the known sides

right you are. Cosine law required.

\[\frac{ \sin52 }{ ? } = \frac{ sinb }{ 4 }\]

\[a^2=b^2+c^2-2bc\cos(A)\]
\[a^2=4^2+7^2-2\times 4\times \cos(52^\circ)\] and a calcullator

hold the phone dear

you do not have three out of any four, because you do not know the angle opposite any known side

@misty1212 i got 5.52507014 from doing your way after square rooting

i can check if you like, i didn't do it

@misty1212 i think it's right but can you check it please

http://www.wolframalpha.com/input/?i=sqrt%284%5E2%2B7%5E2-2*4*7*cos%2852%29%29

5.52 rounded looks good

@misty1212 could you help me with another problem? please

or in your case 5.5 cause they rounded to one decimal place

what do they want, \(r\) ?

@misty1212 trying to find the value of r which is reallyw eird

ok what is the measure of the angle opposite \(r\)?

@misty1212 there's no angle opposite of r that's the weird part

of course there is lol

it is labelled \(\huge R\)

@misty1212 there isn't!!

oh i see, you mean you don't know what it is

lets find it
total in the triangle is 180 so it is
\[180-45-63=72\]

@misty1212 oooh yeah lmao forgot you could do it like that

now law of sines finishes it right?

\[\frac{r}{\sin(72)}=\frac{22}{\sin(45)}\] etc

@misty1212 so it's rearrange to 22sin72=rsin45?

that step is not necessary

\[\frac{r}{\sin(72)}=\frac{22}{\sin(45)}\iff r=22\times\frac{ \sin(72)}{\sin(45)}\]

@misty1212 r=29.58993453?

yeah looks good

@misty1212 thank you!!! you really helped me tonight htis lesson really got me messed up

you're welcome!\[\huge \color\magenta\heartsuit\]