# How do you find the area of triangle ABC given A=10, B=48, C=122, c=11?

Feb 25, 2018

So,the picture looks like this,

So, $\frac{a}{\sin} A = \frac{b}{\sin} B = \frac{c}{\sin} C$

So, $a = c \left(\sin \frac{A}{\sin} B\right) = 2.56$

and, $b = c \left(\sin \frac{B}{\sin} C\right) = 9.64$

So, $s = \frac{11 + 2.56 + 9.64}{2} = 11.6$

So, area of the triangle is = $\sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)} = 11.105$ $u n i {t}^{2}$