Finding angles of triangles

Question

I have what appears to be a $3$-sided triangle: it is two lines on a 180 degree line at the bottom. The bottom left angle is $4x-3$ the top angle is $6x + 3$ and the bottom right angle is not given, but on the angle outside of the triangle between the outside of the triangle and the line is $9x+12$. What am I suppose to do?

Answer 1

Set up a system of equations. You know that the angles of the triangle add up to $180$, and that the angle of a line is also 180. Hence the line at the bottom is equal to $$\theta = 180 - (9x+12).$$

Now you have $$(4x - 3) + (6x +3) + \theta = 180.$$

Solve for $x$, substitute back into each angle measure, and you are finished.

Answer 2

As opposed to those 4-sided triangles? :-)

From your description of the triangle, and the fact that the sum of the interior angles of a triangle is always equal to $180$, we have the equation

$$(4x-3)+(2x-120)+\theta=180,$$

where $\theta$ is the angle we don't know the value of yet.

But, we do know that the supplement of the angle $\theta$ is $9x+12$. Therefore $$\theta=180-(9x+12)=-9x+168.$$ Now plug this into the previous equation and solve for $x$.

Answer 3

I'm including a picture to help "spell things out" visually and for a visual example of "supplementary angles." I've worked the problem out on the image as well.