Shipwreck

(a) The zenith distance is $z=90°-52.5°=37.5°$ and the polar distance is $90°+8.183°=98.18°$. By the sine rule in the astronomical triangle, the angle at P satisfies

so $H^{\prime }=35.557°$. Rigel is high in the sky, so this is the acute value, as the cosine rule confirms once the latitude is known in (d).

The azimuth of 109° puts Rigel east of the meridian, so its local hour angle is −35.557°, which at 15° to the hour is −2h 22m 13.7s, or 21h 37m 46.3s.

(b) 01:00 on 21 November in Bangkok is 18:00 UT on 20 November, 323 days after 1 January. In 323 days the sidereal time at 0h UT gains $323\times 24\text{h}/365.2422=21\text{h}13\text{m}27.4\text{s}$, so at 0h UT on 20 November it is $6\text{h}43\text{m}+21\text{h}13\text{m}27.4\text{s}=3\text{h}56\text{m}27.4\text{s}$, after taking away 24h.

In the next 18 hours of ordinary time it advances by $18\times 1.002738=18\text{h}02\text{m}57.4\text{s}$, so the Greenwich sidereal time is 21h 59m 24.8s, or 21h 59.4m.

(c) The local sidereal time, after taking away 24h, is

The longitude east of Greenwich is the local sidereal time minus the Greenwich sidereal time, $2\text{h}52\text{m}46.3\text{s}-21\text{h}59\text{m}24.8\text{s}=4\text{h}53\text{m}21.5\text{s}$ after adding 24h, which is 73.34° E.

(d) The cosine rule for the side PX, as in (1.20), gives $\sin \delta =\sin \phi \sin h+\cos \phi \cos h\cos A$, which here is

An expression $a\sin \phi -b\cos \phi$ can be written as $R\sin (\phi -\psi )$, with $R=\sqrt{a^2+b^2}$ and $\tan \psi =b/a$. Here $R=\sqrt{{0.7934}^2+{0.1982}^2}=0.8177$ and $\tan \psi =0.1982/0.7934$, so $\psi =14.03°$ and the right-hand side is $0.8177\sin (\phi -14.03°)$, so $\sin (\phi -14.03°)=-0.1741$, which gives $\phi -14.03°=-10.02°$ and $\phi =4.00°\text{ N}$. The other solution of the sine, $\phi -14.03°=190.02°$, is not a latitude.

As a check, (1.19) rearranged with $\phi =4.00°$ gives $\cos H=0.8135$, so the angle at P is 35.56°, as in (a).