Triangle relationships

Question

I was wondering if someone can help me actually. You see I came upon this book called Mathematics for Physics by Michael and Malcolm Woolfson. I a presently stuck on the very first exercise and I can 't seem to find an answer to the question. Unfortunately there isn't a segment showing the workings so I am left to wonder how the hell do I solve this? Nevertheless it has lead me through a multitude of areas that explore the subject matter and in spite of not answering the question yet, I have learned a lot.

But I digress...

The Question goes "Find the side and angles of a triangle with a = 5, b = 6, B = 50 degrees

Just to clarify a few things; this here is a scalene triangle that we are trying to resolve. I will describe is as Triangle ABC (Representing the three angles of the triangles) with a pependicular CP from apex C to side AB (otherwise known as the adjacent). The sides are of lengths a, b and c and the perpendicular is of length h.

I have been given the formula a/sin A = b/sinB = c/sin C

How would you go about solving this?

Answer 1

Given a, b and B, we can obtain A using

a/sin A = b/sinB

Given A and B, we can obtain C (how?).

Given C, we can obtain c using either

b/sinB = c/sin C

or

a/sin A = c/sin C

Answer 2

You have $a, b$ and $B$. You know that $$ \frac{a}{\sin A} = \frac{b}{\sin B}$$ and in that equation three of the quantities are known, so you can solve for $\sin A = \frac56 \sin 50 $.

Now you can use a calculator or whatever to find $A$ by using the inverse sine function. Having $A$ and $B$ you then use your knowledge that the sum of the angles in a triangle is 180 to find $C$. Then use the law of sines one last time$$ \frac{c}{\sin C} = \frac{b}{\sin B}$$ to find $c$.