Answer
$135.7^{\circ}$
Work Step by Step
Given-
$\csc\theta$ = $1.4325$ , $\theta$ belongs to Q II and $0^{\circ}\leq \theta\lt 360^{\circ}$
We will find reference angle of $\theta$ first, using calculator in degree mode by the$\frac{1}{x}$ and $\sin^{-1}$ key with the value $1.4325$ as $\sin\theta$ = $\frac{1}{\csc\theta}$
$1.4325$ → $\frac{1}{x}$ → $\sin^{-1}$→→ 44.273186733
i.e. reference angle of $\theta$ = $44.3^{\circ}$ (to the nearest tenth of a degree)
The desired angle $\theta$ is in Q II, whose reference angle is $44.3^{\circ}$.
Also $0^{\circ}\leq \theta\lt 360^{\circ}$
Therefore-
$\theta$ = $180^{\circ} - 44.3^{\circ}$
= $135.7^{\circ}$
