In the figure above, the line segments meet at a point. If the point

We know
6a+3b = 360
2a+b = 120

Only Option C is 2a+b (RQU)

Hence C

Here we have nine angles. If we cluster them like this way (b + 2a) + (b + 2a) + (b + 2a) we can notice that b + 2a equals 1/3 of 360. RQU is exactly b + 2a. So option C.

If stuck, start with what you know.
The sum of all angles around a point is 360°. (The angles make a circle = 360°) How many angles = \(a\)? How many angles = \(b\)? Count each kind and sum. \(6a +3b=360°\)

Now what? With one equation and two variables, we cannot solve for either variable. Thus, our "measurable angle" will be defined by both \(a\) and \(b\).

Because our answer will contain both variables, we need some (_a + _b) expression. The original equation is our only source. It must be amenable to manipulation. It is. Divide the equation
\(6a +3b=360°\) by \(3\)
\(2a+b=120°\)

So we need an angle composed of one \(b\) and two \(a\)'s. Answer choices require us to start at R. Moving clockwise from R, we do not have much choice.

The first angle between R and S = \(b\)
The next angle between S and T = \(a\)
The next angle between T and U = \(a\)
Angle RQU = \(2a+b=120°\)

Answer C

You can check the other answers if uncertain. The angles in A, B, D, and E respectively are composed of (b), (a+b), (3a+b), and (3a+2b) -- NOT (2a+b).