Question
$${\sin ^{ - 1}}\left\{ {\frac{1}{i}\left( {z - 1} \right)} \right\},$$ where $$z$$ is non-real, can be the angle of a triangle if
A.
$${\text{Re}}\left( z \right) = 1,\operatorname{Im} \left( z \right) = 2$$
B.
$${\text{Re}}\left( z \right) = 1, - 1 \leqslant \operatorname{Im} \left( z \right) \leqslant 1$$
C.
$${\text{Re}}\left( z \right) + \operatorname{Im} \left( z \right) = 0$$
D.
None of these
Answer :
$${\text{Re}}\left( z \right) = 1, - 1 \leqslant \operatorname{Im} \left( z \right) \leqslant 1$$
Solution :
$$\frac{{z - 1}}{i}$$ must be real. Now, $$\frac{{z - 1}}{i} = \frac{{x - 1 + iy}}{i} = y - \left( {x - 1} \right)i$$
$$\therefore \,\,x - 1 = 0.$$ Then $${\sin ^{ - 1}}\left\{ {\frac{1}{i}\left( {z - 1} \right)} \right\} = {\sin ^{ - 1}}y.\,{\text{So, }} - 1 \leqslant y \leqslant 1.$$