A man travel from Pekin (ΦP = 39◦54′N, ΛP = 116◦23′E) to Singapur (ΦS = 1◦17′N, ΛS = 103◦51′E). Suppose the trajectory is a great circle:
a) find the distance in nmi betwen Pekin and Singapur
b) find the final rhumb in Singapur
c) find the distance from Singapur to the point X ( intersection betwten the trajectory Pekin-Singapur and the equator.
Answers
a)
$\text{N}=\Lambda_P-\Lambda_S=116º23'-103º51'=12º32'$
$p=90º-\Phi_S=90º-1º17'= 88º43'$
$s=90º-\phi_P =90º-39º54'=50º6'$
Using the cosine rule:
$$\cos n = \cos s \cos p + \sin s \sin p \cos \text{N}$$
$$n=\arccos (\cos s \cos p + \sin s \sin p \cos \text{N})$$
$ n = \arccos( \cos (50º6')\cos (88º43') + \sin (50º6') \sin (88º43')\cos (12º32')$ \
$$\boldsymbol{n}= 40º15'54.43'' = \boldsymbol{2415.90717}\, \textbf{nmi}$$
Is this correct?
For b) I don't know excatly how solve it, but I did this:
The final rhumb is $180º+S$. $$\frac{\sin N}{\sin n} = \frac{\sin S}{\sin s}\Rightarrow \sin S= \frac{\sin N\sin s}{\sin n}\Rightarrow S = \arcsin\left( \frac{\sin N\sin s}{\sin n}\right)$$
$$ S= \arcsin\left( \frac{\sin (12º32')\sin(50º6') }{\sin (40º15'54.43'')}\right) = 14º55'35.51''$$
$$ \textbf{Final rhumb}=180º+14º55'35.51''= \boldsymbol{ 194º 55'35.51''} =\textbf{S}\boldsymbol{14.9265303º}\, \textbf{W}$$
For c what I did is:
First made a new spherical triangle with points $N,X,S'$. We know that
$p = 88º43', s'=90º, S'$ =180-S=$ 165º 4'24.49''$
$$\cos s' =0= \cos n' \cos p + \sin n' \sin p \cos S' $$ $$ \cos n'\cos p = -\sin n'\sin p \cos S'$$ $$ \cos n' = -\sin n'\tan p \cos S '$$ $$ \cot n' = -\tan p \cos S'$$ $$n'=\arctan\left( (-\tan p \cos S'\right)^{-1})$$ $$\boldsymbol{n'}=1º19'41.28''=\boldsymbol{79.68800445}\,\textbf{nmi}$$
Can you help me please? I don't know if my answers are correct
I did a draw with Geogebra but I don't know if it is right ...