Question: If one angle of a parallelogram is $24^{\circ}$ less than twice the smallest angle, then what is the measure of the largest angle of the parallelogram?
My attempt:
First of all, I drew a picture.
Now, there are two choices for selecting the first angle of the parallelogram:
- $\angle A$ (or $\angle C$)
- $\angle B$ (or $\angle D$)
Suppose we start with $\angle A$ (or $\angle C$).
According to the question,
$$\angle A = 2\angle B - 24^{\circ}$$
Since $\angle A$ and $\angle B$ form co-interior angles, their sum will be $180^\circ$.
$$\begin{align} &180^\circ - \angle B = 2 \angle B - 24^\circ \\ \implies & 180^\circ + 24^\circ = 3 \angle B \\ \implies &\angle B = \frac{204}{3} = 68^\circ \end{align}$$ Therefore, the largest angle is $180^\circ - 68^\circ = 112^\circ$.
If we start with another angle $\angle B$ (Or $\angle D$) we have, $$\begin{align}\angle B &= 2\angle B - 24^\circ \\ \implies \angle B &= 24^\circ \end{align} $$ This result corresponds to the smallest angle.
So the largest angle is $180^\circ - 24^\circ = 156^\circ$
Therefore the final answer should be $156^\circ$.
Is this correct? According to my book, the answer should be $112^\circ$. It doesn't consider the another case.