Question

Solve the triangle using the Law of Sines. (Assume c = 55, _A = 549, and zB = 20°. Round lengths to two decimal places.) I just want a....and b

Answer 1

Saw of Lines:

`(a)/(\sin A)=(b)/(\sin B)=(c)/(\sin C)`

Side c is given as

c = 55

Also,

b = 20°

Now, we can use the law and solve. Shown below:

`\begin{gathered} (b)/(\sin B)=(c)/(\sin C) \\ (b)/(\sin20)=(55)/(\sin 106) \\ b\sin 106=55\sin 20 \\ b=(55\sin 20)/(\sin 106) \\ b=19.57 \end{gathered}`

Then, we want to solve for a. Shown below:

`(a)/(\sin A)=(c)/(\sin C)`

So, substituting, we have:

`\begin{gathered} (a)/(\sin 54)=(55)/(\sin 106) \\ a\sin 106=55\sin 54 \\ a=(55\sin 54)/(\sin 106) \\ a=46.29 \end{gathered}`