1. ## trig help

B=5 deg C=125 deg b=200
round to the nearest tenths and lengths

2. 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

3. Hello, Shorty!

I assume the instructions are "Solve the triangle".

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

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

I prefer to organize the information in a chart.

. . $\begin{array}{ccc}A = \\B = \\ C = \end{array}
\begin{array}{ccc}a = \\ b = \\ c = \end{array}

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

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

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

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

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

. . $\boxed{\begin{array}{ccc}A = \\B = \\ C = \end{array}