\(M P=\frac{\tan \beta( \text { angle subtended by image at eye piece) } }{\tan \alpha \text { (angle subtended by object on objective) }}\)

Also,  \(M P =\frac{f_{o}}{f_{e}}=\frac{150}{5}=30\)

\(\tan \alpha =\frac{50}{1000}=\frac{1}{20}\, \mathrm{rad}\)

\(\therefore \quad \tan \beta=\theta=M P \times \tan \alpha\)

\(=30 \times \frac{1}{20}=\frac{3}{2}=1.5\)

or, closest value of \(\beta \approx 60^o\)