Question

${\sin }^{-1}\left\{\frac{1}{i}\left(z-1\right)\right\}$, where $z$ is nonreal, can be the angle of a triangle if

• A. $\mathrm{Re}\left(z\right)=1,\mathrm{Im}\left(z\right)=2$
• B. $\mathrm{Re}\left(z\right)=1,0<\mathrm{Im}\left(z\right)\le 1$
• C. $\mathrm{Re}\left(z\right)+\mathrm{Im}\left(z\right)=0$
• D. $\mathrm{Re}\left(z\right)=2,0<\mathrm{Im}\left(z\right)\le 1$

Answer 1

Answer: (B)

$\mathrm{Re}\left(z\right)=1,0<\mathrm{Im}\left(z\right)\le 1$

${\sin }^{-1}\left(\frac{1(z-1)}{i}\right)Tomakethisreal,-1\le \frac{1(z-1)}{i}\le 1andtomakethisastriangl{e}^{\prime }sangle0<\frac{1(z-1)}{i}\le 1Now,forthistoexist(z-1)mustbepurelyimaginaryThereforeRe\left(z\right)=1andforthistoliebetween0and10<Im\left(z\right)\le 1$