If $\sin(a) = 1 / \sqrt 5$ and $\sin(b) = 3/5$, then $b - a$ lies in which interval?

Question

If $\sin(a) = \dfrac{1}{\sqrt 5}$ and $\sin(b) = \dfrac35$, then $b - a$ lies in which of the following intervals?

a) $\left[0, \dfrac{\pi}{4}\right]$

b) $\left[\dfrac\pi2, \dfrac{3\pi}{4}\right]$

c) $\left[\dfrac{3\pi}{4}, \pi\right]$

d) $\left[\pi, \dfrac{5\pi}{4}\right]$

Please explain.

Answer 1

I suspect you are not allowed calculators and the intended answer is a. You are expected to know and reason that $\sin x$ is increasing on $[0,\frac \pi 2]$ and that $\sin \frac \pi 4 = \frac 1{\sqrt 2} \approx 0.707 \gt \frac 35$, so $b \lt \frac \pi 4$ and $b-a \in [0,\frac \pi 4]$. sharding4 has shown that the question is defective as b could be an answer as well.

Answer 2

I believe without some additional information about the quadrants in which $a$ and $b$ terminate, the question is defective and either choice a) or choice b) would be possible. You can work numerically in degrees. $a$ is either $26.6^{\,\circ}$ or $153.4^{\,\circ}$ and $b$ is $36.9^{\,\circ}$ or $143.1^{\,\circ}$ What are the possible values for $b-a$ and how do those compare to your choices?