# trig help

• Nov 11th 2006, 03:07 AM
SHORTY
trig help
B=5 deg C=125 deg b=200
round to the nearest tenths and lengths
• Nov 11th 2006, 04:00 AM
topsquark
Quote:

Originally Posted by SHORTY
B=5 deg C=125 deg b=200
round to the nearest tenths and lengths

And what exactly do you want us to do with this information? Ask us a question!

-Dan
• Nov 11th 2006, 04:38 AM
Soroban
Hello, Shorty!

I assume the instructions are "Solve the triangle".

You're expected to be familiar with the Law of SInes:
. . $\displaystyle \frac{a}{\sin A} = \frac{b}{\sin b} = \frac{c}{\sin C}$

Quote:

$\displaystyle B = 5^o,\;C = 125^o,\;b = 200$
Round to the nearest tenth.

I prefer to organize the information in a chart.

. . $\displaystyle \begin{array}{ccc}A = \\B = \\ C = \end{array} \begin{array}{ccc}\boxed{\quad} \\ 5^o \\ 125^o\end{array}\quad \begin{array}{ccc}a = \\ b = \\ c = \end{array} \begin{array}{ccc}\boxed{\quad} \\ 200 \\ \boxed{\quad}\end{array}$

Since the angles total 180°, we have: $\displaystyle A \:= \:180^o - 5^o - 125^o \:= \:50^o$
Hence, we have:
. . $\displaystyle \begin{array}{ccc}A = \\B = \\ C = \end{array} \begin{array}{ccc}\boxed{50^o} \\ 5^o \\ 125^o\end{array}\quad \begin{array}{ccc}a = \\ b = \\ c = \end{array} \begin{array}{ccc}\boxed{\quad} \\ 200 \\ \boxed{\quad}\end{array}$

To find side $\displaystyle c$, use: .$\displaystyle \frac{c}{\sin C} = \frac{b}{\sin B}\quad\Rightarrow\quad c \:=\:\frac{b\sin C}{\sin B}$

. . and we have: .$\displaystyle c\:=\:\frac{200\sin125^o}{\sin5^o} \:\approx\:1879.7$

To find side $\displaystyle a$, use: .$\displaystyle \frac{a}{\sin A} = \frac{b}{\sin B}\quad\Rightarrow\quad a \:=\:\frac{b\sin A}{\sin B}$

. . and we have: .$\displaystyle a \:=\:\frac{200\sin50^o}{\sin5^o} \:\approx\:1757.9$

. . $\displaystyle \boxed{\begin{array}{ccc}A = \\B = \\ C = \end{array} \begin{array}{ccc}\boxed{50^o} \\ 5^o \\ 125^o\end{array}\quad \begin{array}{ccc}a = \\ b = \\ c = \end{array} \begin{array}{ccc}\boxed{1757.9} \\ 200 \\ \boxed{1879.7}\end{array}}$ . . . There!