Question

Find the reference angle for ${{108}^{\circ }}$.
A. $\rho ={{50}^{\circ }}$
B. $\rho ={{60}^{\circ }}$
C. $\rho ={{72}^{\circ }}$
D. $\rho ={{40}^{\circ }}$

Answer 1

Hint: We first try to establish the concept of reference angle. We find out the relation between two reference angles in different conditions. We try to find out the position of the given angle. Then according to the conditions, we find out the reference angle for ${{108}^{\circ }}$

Complete step-by-step answer:
We try to break the angles inside ${{360}^{\circ }}$ into the four normal quadrants. There are 4 quadrants - first, second, third, fourth. Now for any kind of given angle $\alpha $, the reference angle $\rho $ will be the acute angle that the terminal side of the angle $\alpha $ makes with the x-axis.
These angles generally have non-quadrantal angles in normal positions. We always need to find the difference between the angle and the positive or negative x-axis.
Now we express the quadrant wise different results.
For given angle $\alpha $ in the first quadrant $\left( {{0}^{\circ }}<\alpha <{{90}^{\circ }} \right)$, the reference angle will be $\rho =\alpha $.
For given angle $\alpha $ in the second quadrant $\left( {{90}^{\circ }}<\alpha <{{180}^{\circ }} \right)$, the reference angle will be $\rho ={{180}^{\circ }}-\alpha =\pi -\alpha $.
For given angle $\alpha $ in the third quadrant $\left( {{180}^{\circ }}<\alpha <{{270}^{\circ }} \right)$, the reference angle will be $\rho =\alpha -{{180}^{\circ }}=\alpha -\pi $.
For given angle $\alpha $ in the first quadrant $\left( {{270}^{\circ }}<\alpha <{{360}^{\circ }} \right)$, the reference angle will be $\rho ={{360}^{\circ }}-\alpha =2\pi -\alpha $.
The representation will be of the form

For the given problem we need to find a reference angle for ${{108}^{\circ }}$.
The angle ${{108}^{\circ }}$ is in the second quadrant as $\left( {{90}^{\circ }}<{{108}^{\circ }}<{{180}^{\circ }} \right)$.
So, using the formulas the reference angle will be $\pi -{{108}^{\circ }}={{180}^{\circ }}-{{108}^{\circ }}={{72}^{\circ }}$.

So, the correct answer is “Option C”.

Note: We don't perfectly memorise the given formulas for all four quadrants. We only need to remember that the reference angle will be the acute angle that the terminal side of the given angle makes with the x-axis. Using this logic, we can ourselves find both the formula and the solution.