Find the latitude of an observer in the northern hemisphere if it is known, at a certain time, the time angle of the Sun $H_S = 1^h34^m24^s$ , the altitude of the Sun $h_S = 40º10 '$and the declination of the Sun $\delta_S = 15º38'$
Answer:
Using the formula $\sin h_s=\sin\Phi\sin\delta+\cos\delta\cos\Phi\cos H_s $ we have:
$\sin h_s=\sin\Phi\sin\delta+\sqrt{1-\sin^2\Phi}\cos H_s\cos\delta\Rightarrow \sin h_s-\sin\Phi\sin\delta=\sqrt{1-\sin^2\Phi}\cos H_s\cos\delta\Rightarrow(\sin h_s-\sin\Phi\sin\delta)^2=(1-\sin^2\Phi)\cos^2H_s\cos\delta=\cos^2 H_s\cos^2\delta-\cos^2 H_s\cos^2\delta\sin^2\Phi\Rightarrow $ $ \sin h_s^2-2\sin h_s\sin\Phi\sin\delta+\sin^2\Phi\sin^2\delta=\cos^2 H_s\cos^2\delta-\cos^2H_s\cos^2\delta\sin^2\Phi\Rightarrow$ $ \cos^2 H_s\cos^2\delta\sin^2\Phi+\sin^2\Phi\sin^2\delta-2\sin h_s\sin\Phi\sin\delta+\sin h_s^2-\cos^2 H_s=0 \Rightarrow \\(\cos^2 H_s+\sin^2\delta)\sin^2\Phi-2\sin h_s\sin\Phi\sin\delta+\sin h_s^2-\cos^2 H_s\cos^2\delta=0$
Remplacing with the information we have:
$(\cos^223º36'\cos^215º38'+\sin^215º38')\sin^2\Phi-2(40º10'\sin15º38')\sin\Phi+\sin^2(40º10')-\cos^2 23º36'\cos^2 15.38'= 0 \, 0.8513\sin^2\Phi-0.3476\sin\Phi -0.3626=0$
Then solving the cuadratic equation we have:
$1) \sin\Phi=-0.3174\rightarrow \Phi=-28º46'31.81'' \\ 2)\sin \Phi=0.8880556\rightarrow \Phi=62º37'47.72''$
Because the observer is in the northern hemisphere I choose the 2 option.
Is my answer correct?
The answer was correct. This is the fastest way to solve it.