# Determining the Sides of a Triangle, Given are Perimeter and three Angles

I was studying for some exams when a wild question appeared. It looks like this:

Given that the perimeter of a triangle is $180$ inches. If the angles of the triangles are in the ratio $5:6:7$, determine the sides of the triangle.

My work:

Well, the only way that the angles found in the triangle to be in the ratio $5:6:7$ is when the three angles in the triangle were $50$, $60$ and $70$ degrees. I don't know what might be the three angles of the triangle if the ratio of those three becomes arbitrary, say 11:12:5.

I imagine holding a wire with length 180 inches. but I don't know how far from the tip of the wire will I bend to make an angle of 50 degrees and the succeeding angles in triangle.

I'm in a mess. How will you find the sides of the triangle?

• The Law of Sines might help here. – lulu Aug 17 '17 at 13:50

By law of sines: $$\frac{a}{\sin50^{\circ}}=\frac{b}{\sin60^{\circ}}=\frac{c}{\sin70^{\circ}}$$ and $a+b+c=180$.
Let $\frac{a}{\sin50^{\circ}}=k$.
Hence, $a=k\sin50^{\circ}$, $b=k\sin60^{\circ}$, $c=k\sin70^{\circ}$ and $$k\sin50^{\circ}+k\sin60^{\circ}+k\sin70^{\circ}=180,$$ which gives $$k=\frac{180}{\sin50^{\circ}+\sin60^{\circ}+\sin70^{\circ}}$$ and from here we can get a values of $a$, $b$ and $c$.
• Wow! I never thought about writing this: $k=\frac{180}{\sin50^{\circ}+\sin60^{\circ}+\sin70^{\circ}}$. What property in trigonometry did you exploit or used? – Palautot Ka Aug 17 '17 at 14:03