The angle of elevation \( \theta \) of the sun in the sky at any latitude \( L \) is calculated with the formula below. In the formula, \( \theta=0 \) corresponds to sunrise and \( \theta=\frac{\pi}{2} \) occurs if the sun is directly overhead. The greek letter omega, \( \omega \), is the number of radians that Earth has rotated through since noon, when \( \omega=0 \). \( D \) is the declination of the sun, which varies because Earth is on a tilted axis. Town A has a latitude \( L=81.50^{\circ} \), or 1.4224 radian. Find the angle of elevation \( \theta \) of the sun at 10 P.M. on July 4,2012 , where at that time \( D \approx 0.2402 \) and \( \omega \approx 2.6180 \). \( \sin \theta=\cos D \cos L \cos \omega+\sin D \sin L \) \( \theta \approx \square \) (Type your answer in radians. Round to four decimal places as needed.)

To find the angle of elevation \( \theta \) of the sun, we can plug the given values into the formula: \[ \sin \theta = \cos D \cos L \cos \omega + \sin D \sin L \] Given: - \( D \approx 0.2402 \) - \( L \approx 1.4224 \) - \( \omega \approx 2.6180 \) First, we need to calculate each component: 1. \( \cos D \approx \cos(0.2402) \approx 0.9703 \) 2. \( \cos L \approx \cos(1.4224) \approx 0.1564 \) 3. \( \cos \omega \approx \cos(2.6180) \approx -0.8670 \) 4. \( \sin D \approx \sin(0.2402) \approx 0.2397 \) 5. \( \sin L \approx \sin(1.4224) \approx 0.9877 \) Now, substituting into the equation: \[ \sin \theta \approx (0.9703)(0.1564)(-0.8670) + (0.2397)(0.9877) \] Calculating the terms: 1. \( (0.9703)(0.1564)(-0.8670) \approx -0.1281 \) 2. \( (0.2397)(0.9877) \approx 0.2360 \) Combine these: \[ \sin \theta \approx -0.1281 + 0.2360 \approx 0.1079 \] To find \( \theta \), take the inverse sine: \[ \theta \approx \arcsin(0.1079) \approx 0.1080 \text{ radians} \] Thus, the angle of elevation \( \theta \) of the sun at 10 P.M. on July 4, 2012 is approximately: \[ \theta \approx 0.1080 \text{ radians} \]