![\huge\boxed{a=\frac{229}{2};\ \ b=\frac{51}{2}}](https://tex.z-dn.net/?f=%5Chuge%5Cboxed%7Ba%3D%5Cfrac%7B229%7D%7B2%7D%3B%5C%20%5C%20b%3D%5Cfrac%7B51%7D%7B2%7D%7D)
<em>In decimal form, a = 114.5; b = 25.5</em>
We will write this as a system of equations, where
and
are the two angles.
![\left \{ {{a=5b-13} \atop {a+b=140}} \right](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7Ba%3D5b-13%7D%20%5Catop%20%7Ba%2Bb%3D140%7D%7D%20%5Cright)
We will use substitution to solve this system. We know what
equals, so we plug that into the second equation.
![5b-13+b=140](https://tex.z-dn.net/?f=5b-13%2Bb%3D140)
Combine the like terms.
![6b-13=140](https://tex.z-dn.net/?f=6b-13%3D140)
Add
on both sides.
![6b=153](https://tex.z-dn.net/?f=6b%3D153)
Divide both sides by
.
![b=\frac{153}{6}=\boxed{\frac{51}{2}}](https://tex.z-dn.net/?f=b%3D%5Cfrac%7B153%7D%7B6%7D%3D%5Cboxed%7B%5Cfrac%7B51%7D%7B2%7D%7D)
Now just subtract that from
.
![a=140-\frac{51}{2}=\boxed{\frac{229}{2}}](https://tex.z-dn.net/?f=a%3D140-%5Cfrac%7B51%7D%7B2%7D%3D%5Cboxed%7B%5Cfrac%7B229%7D%7B2%7D%7D)
D because d is always right hahahahaha
If you look carefully at the graph, you may see that the slope of the line is
3-4 -1
---------------- = ------ = m
7-4 3
thus, you have the slope of the line and two points on the line. Suppose we
choose the point (4,4) and subst. the known slope and the coordinates of this point into the point-slope formula for the eqn of a str line:
y-y1 = m (x-x1)
y-4 = (-1/3)(x-4)
This is the desired equation. You could, if you wished, change this into slope-intercept form.