## Q. 7.3

Where Is the Sun? Find the altitude angle and azimuth angle for the sun at 3:00 P.M. solar time in Boulder, Colorado (latitude 40º) on the summer solstice.

## Verified Solution

Since it is the solstice we know, without computing, that the solar declination δ is 23.45º. Since 3:00 P.M. is three hours after solar noon, from (7.10) we obtain

$H \ = \ \left(\frac{15º}{h}\right) \ \cdot \ \left(\text{hours before solar noon}\right) \ = \ \frac{15º}{h} \ \cdot \ \left(-3 \ h\right) \ = \ −45º$

Using (7.8), the altitude angle is

$\begin{matrix} \sin \ \beta & = \ \cos \ L \ \cos \ \delta \ \cos \ H \ + \ \sin \ L \ \sin \ \delta \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ \\ & = \ \cos \ 40º \ \cos \ 23.45º \ \cos \left(-45º\right) \ + \ \sin \ 40º \ \sin \ 23.45º \ = \ 0.7527 \\ \quad \ \ \beta & = \ \sin^{-1} \left(0.7527\right) \ = \ 48.8º \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \ \ \ \end{matrix}$

From (7.9) the sine of the azimuth angle is

$\begin{matrix} \sin \ \phi _{S} & = \ \frac{\cos \ \delta \ \sin \ H}{\cos \ \beta} \quad \quad \quad \quad \quad \quad \quad \quad \\ & = \ \frac{\cos \ 23.45º \ \sin \ \left(-45º\right)}{\cos \ 48.8º} \ = \ −0.9848 \end{matrix}$

But the arcsine is ambiguous and two possibilities exist:

$\begin{matrix} \phi_{S} & = \ \sin^{-1} \left(−0.9848\right) \ = \ −80º \quad \left(80º \text{west of south}\right) \ \ \ \ \ \ \\ \text{or} & \phi_{S} \ = \ 180 \ – \ \left(-80\right) \ = \ 260º \quad \quad \left(100º \text{west of south}\right) \end{matrix}$

To decide which of these two options is correct, we apply (7.11):

$\text{if} \quad \cos \ H \ \geq \ \frac{\tan \ \delta}{\tan \ L}, \quad \text{then} \ \left|\phi_{S}\right| \ \leq \ 90º; \quad \text{otherwise} \ \left|\phi_{S}\right| \ \gt \ 90º$ (7.11)
$\cos \ H \ = \ \cos \left(-45º\right) \ = \ 0.707 \quad \text{and} \quad \frac{\tan \ \delta}{\tan \ L} \ = \ \frac{\tan \ 23.45º}{\tan \ 40º} \ = \ 0.517$

Since $\cos \ H \ \geq \ \frac{\tan \ \delta}{\tan \ L}$ we conclude that the azimuth angle is

$\phi_{S} \ = \ -80º \quad \left(80º \text{west of south}\right)$