Find the length which at a distance of \(5280\) m will subtend an angle of \(1’\) at the eye.
Solution:
We know:
\[ s=r\theta \]
Given:
\[ r=5280 \text{ m} \]
\[ \theta=1′ \]
Since,
\[ 1^\circ=60′ \]
\[ 1’=\frac{1}{60}^\circ \]
Convert into radians:
\[ \theta=\frac{1}{60}\times\frac{\pi}{180} = \frac{\pi}{10800} \]
Now,
\[ s=5280\times\frac{\pi}{10800} \]
\[ s=\frac{22\pi}{45} \]
\[ s\approx1.54 \text{ m} \]
Therefore, the required length is:
\[ \boxed{1.54 \text{ m (approximately)}} \]