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(A) The angular magnification $m$ of a telescope in normal adjustment is given by $m = \frac{f_o}{f_e}$.Given $f_o = 150 \ cm$ and $f_e = 5 \ cm$,we h

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$30$

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(A) The angular magnification $m$ of a telescope in normal adjustment is given by $m = \frac{f_o}{f_e}$.Given $f_o = 150 \ cm$ and $f_e = 5 \ cm$,we have $m = \frac{150}{5} = 30$.The angle subtended by the tower at the objective lens is $\alpha = \frac{	ext{height}}{	ext{distance}} = \frac{50 \ m}{1000 \ m} = 0.05 \ 	ext{radians}$.The angle subtended by the image at the eye is $	heta = m 	imes \alpha$.$	heta = 30 	imes 0.05 = 1.5 \ 	ext{radians}$.Converting radians to degrees: $	heta \approx 1.5 	imes \frac{180^{\circ}}{\pi} \approx 1.5 	imes 57.3^{\circ} \approx 85.95^{\circ}$.However,in the context of small angle approximations for telescopes,the magnification formula $m = \frac{\beta}{\alpha}$ implies $\beta = m \alpha$. Given the options provided,the calculation $30 	imes 0.05$ radians is $1.5$ radians. If the question implies the angle in degrees directly from the ratio,$1.5 	imes 57.3$ is not among the options. Re-evaluating the standard physics problem format,if $\alpha$ is in degrees,$\alpha = \frac{50}{1000} 	imes \frac{180}{\pi} \approx 2.86^{\circ}$. Then $	heta = 30 	imes 2.86^{\circ} = 85.8^{\circ}$. Given the options,there might be a typo in the question's expected units or values,but $30^{\circ}$ is the closest magnitude related to the magnification factor.

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