Spherical Trigonometry Solutions

 Problems:

1) Show that: 

a) (i^ x j^)  =  sin c m =  sin c (⊥ i   +  j)

b) (i^ x k^)  =  sin b u = sin b (⊥i^ + k^)

Solutions:  

(1)In each case (a and b) we are dealing with the vector cross product.  This is illustrated for the simpler case below:

Here, we let the angle subtended, e.g. between A and B be Θ  with:  0   <  Θ  <  π.  Then unless A and B are parallel, they now determine a plane.  Let n be a unit vector perpendicular to the plane and pointing in the direction a right-handed thread screw would advance when its head is rotated from A to B through the angle Θ .

The vector product or cross product is then:

A X B    =   n  ‖A‖ ‖B‖   sin  Θ

If we now examine the spherical triangle diagram (Oct. 31 post) with the vectors (i^ x j^)  and  (i^ x  k^)   :

We see  (i^ x j^)  subtends the angle c  and the cross product  (i^ x k^) subtends the angle b.  Hence, by the same principle as in the simplified case:

(i^ x j^)  =  sin c m= sin c (⊥ i   +  j)

(i^ x k^)  =  sin b u = sin b (⊥ i^ + k^)

Where u = (⊥ i   +  k),  m = (⊥ i   +  j) are the orientation vectors

2) In a spherical triangle ABC, C  =  90 ^o  , a = 119 ^o  30'  and B =  52 ^o.5.  Calculate the values of b, c and A.

Solution:

Since  C  =  90 ^o  we have a right-angled spherical triangle. Hence, we need:  tan b  =  sin a tan B

B =  52 ^o.5    and  a = 119 ^o  30'  = a = 119 ^o .5 

Using consistent decimal degree units.

Then:  tan b = sin (119 ^o .5) tan ( 52 ^o.5 )

From a trig table (or calculator):

sin (119 ^o .5) =  0. 87035

tan ( 52 ^o.5 )  =  1.30323

Then:  tan b = (0.87035) (1.30323)

tan b = 1.1341

And:  arctan (1.1341) = b  =  48 ^o  36'

To find c we can use part of the fundamental formula, relating c to a and b:

 

cos c =  cos a cos b

cos c =  cos (119 ^o.5) cos (48 ^o.6)

cos c =  (- 0.4924) (0.6613) =  - 0.3256

Then: c = arc cos(-0.3256) = 109 ^o.02

To lastly find angle A, we can use that part of the law of sines relating A to angle B and sides a, c,  i.e.

sin A =  sin a sin B/ sin c

sin A =  sin (119 ^o .5) sin ( 52 ^o.5 )/ sin (109 ^o.02)

sin A = (0.6903)/ (0.9453) =  0.7302

arc sin (0.7302) = A  =  46 ^o.9   Or:   A = 46 ^o  54'

3)The altitude of a star as it transits your meridian is found to be 45^o along a vertical circle at azimuth 180^o, the south point.  Find the declination (d ) of the star.

Solution:

From the celestial sphere geometry:

Here  a _max   = 45^o    is the altitude at meridian transit at the (unspecified) observer location (hence a maximum). 

For a star transiting the meridian south of the zenith (ref. Aug. 1 post):

Declination = Observer's latitude -  Meridian z. d. (z')

 

Or:  d  =  f(latitude) -  z'

Since:   90^o -  a _max   =   z'

d  =  f(latitude) -  ( 90^o -  a _max )

Then:

d  =  f(latitude) -  ( 90^o -  45^o )

Þ  d  =  f   -   45^o  

For an observer at a latitude of  45^o  N (e.g. just north of Green Bay, WI) we have:

d  =   45^o   -   45^{o   =}   0 ^o

I.e. the star is on the celestial equator