grr.... so fustrated!!!!

[horrible_at_tests]

:evil: i hate math so much.... i have been trying to complete this one question for a while...
SinA =SinB=SinC
a = b = c

what the question has for numbers are:
SinA=Sin27=SinC
a = 15 = 18


so i no that i can use SinB and Sin C since those are the only two that have numbers in them

so one i do that i cross multiply so then i will get

18Sin27=15SinC


after that im not sure what to do because i no that i have to get C by itself but its kinda tough because im not sure if i have to divide the 15Sin by the 18Sin27 or what ... grr... its so fustrating

 

[soroban]

Hello, horrible_at_tests!

\(\displaystyle \text{Given: }\;B = 27^o,\;b = 15,\;c = 18,\;\;\text{solve the triangle.}\)

I organize the information in a chart . . .

\(\displaystyle \begin{array}{ccccccc}A &= &\boxed{\quad} & \quad & a & = & \boxed{\quad} \\ B & = & 27^o & \quad & b & = & 15 \\ C & = & \boxed{\quad} & \quad & c & = & 18 \end{array}\)


\(\displaystyle \text{To find angle }C\text{, we use: }\;\frac{\sin C}{c} \:=\:\frac{\sin B}{b}\)

\(\displaystyle \text{and we have: }\;\frac{\sin C}{18} \:=\:\frac{\sin27^o}{15}\quad\Rightarrow\quad\sin C \:=\:\frac{18\!\cdot\!\sin27^o}{15} \:=\:0.5447886\)

. . \(\displaystyle \text{Therefore: }\;\boxed{C \:\approx\:33.0^o}\)


\(\displaystyle \text{Then: }\;A \:=\:180^o - 27^o - 33^o \quad\Rightarrow\quad\boxed{ A\:=\:120^o}\)


\(\displaystyle \text{We have: }\;\frac{a}{\sin A} \:=\:\frac{b}{\sin B}\)

\(\displaystyle \text{Hence: }\;\frac{a}{\sin120^o} \:=\:\frac{15}{\sin27^o}\quad\Rightarrow\quad a \:=\:\frac{15\!\cdot\!\sin120^o}{\sin27^o} \:=\:28.6137729\)

. . \(\displaystyle \text{Therefore: }\;\boxed{a \:\approx\:28.6}\)

 

[galactus]

:evil: i hate math so much.[/quote

[quote:32z2gymr]so i no that i can use SinB and Sin C since those are the only two that have numbers in them

so when i do that i cross multiply so then i will get

18Sin27=15SinC

[/quote:32z2gymr]

Apparently, you hate English as well :wink:

No, you can not divide by Sin.

If you have \(\displaystyle 18sin(27)=15sin(C)\)

Divide by 15 and then use arcsine.

\(\displaystyle \frac{18sin(27)}{15}=sin(C)\)

\(\displaystyle C=sin^{-1}(\frac{18sin(27)}{15})\)

 

[Loren]

galactus

"Cannot" is one word.

 

[stapel]

"Cannot" is one word.

Actually, each is acceptable, according to the generally-recognized authority on the English language. :wink:

Eliz.

 

[galactus]

galactus

"Cannot" is one word.

People in glass houses, right?. I just had an inadvertent space. I hate to see posts written in 'IM-speak'.

i.e. 'no', when 'know' is what is really meant, 'u' for 'you', etc.