# How do you find the area of triangle ABC given A=171^circ, B=1^circ, C=8^circ, b=2?

Nov 4, 2016

Assuming that I didn't make any calculator mistakes:
color(white)("XXX")"Area"_triangle~~2.494953309 sq.units

#### Explanation: By the Law of Sines:
$\textcolor{w h i t e}{\text{XXX}} \frac{a}{\sin \left(A\right)} = \frac{b}{\sin \left(B\right)} = \frac{c}{\sin \left(C\right)}$

For the given values this means:
$\textcolor{w h i t e}{\text{XXX}} a = \frac{\sin \left({171}^{\circ}\right) \cdot 2}{\sin \left({1}^{\circ}\right)}$

$\textcolor{w h i t e}{\text{XXXx}} \approx 17.92697937$

and
color(white)("XXX")c=(sin(8^circ)*2)/(sin(1^circ)

$\textcolor{w h i t e}{\text{XXXx}} \approx 15.94887232$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

This gives a perimeter of
$\textcolor{w h i t e}{\text{XXX}} p = 17.92697937 + 2 + 15.94887232$
$\textcolor{w h i t e}{\text{XXXx}} = 35.87585168$
and a semi-perimeter of
$\textcolor{w h i t e}{\text{XXX}} s = 35.87585168 \div 2$
$\textcolor{w h i t e}{\text{XXXx}} = 17.93792584$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Using Heron's Formula for the Area of a Triangle
color(white)("XXX")"Area"_triangle=sqrt(s(s-a)(s-b)(s-c))

$\textcolor{w h i t e}{\text{XXXXXXX}} \approx 2.494953309$