Question

Consider a triangle like the one below. Suppose that A=61 degrees,C=63 degrees , and a=55.

Find B(the angle)
b(the side)
c(the side)

Answer 1

Check the picture below.

Make sure your calculator is in Degree mode.

`\textit{Law of sines} \\\\ \cfrac{\sin(\measuredangle A)}{a}=\cfrac{\sin(\measuredangle B)}{b}=\cfrac{\sin(\measuredangle C)}{c} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \cfrac{b}{\sin(56^o)}=\cfrac{55}{\sin(61^o)}\implies b=\cfrac{55\sin(56^o)}{\sin(61^o)}\implies b\approx 52.1 \\\\\\ \cfrac{c}{\sin(63^o)}=\cfrac{55}{\sin(61^o)}\implies c=\cfrac{55\sin(63^o)}{\sin(61^o)}\implies c\approx 56.0`