Question

SPHERICAL TRIGONOMETRY

1. solve for the three vertices of a spherical triangle ABC if the sides measures a=68degree48', b=53degree15' and c=46degree30'.
2. solve for these right spherical triangle with parts side c=69degree25' and angle B=63degree25' note that this is a right spherical triangle. (HINT: use the Napier's circle to determine what can be solved first by the given parameters.)
3. solve for the area of the spherical triangle in problem 1 and 2 if the sphere has a radius of 30m.
4. *show solution*

Answer 1

compilation of the THREE questions

(1) we shall solve the first question using the Haversine formula

$havA=\frac{hav \hspace{0.1cm} a - hav(b-c)}{sin b* sin c}$

We convert the give parameters to degrees

$a=68°41^{\prime}=68.8°\\b=54°15^{\prime}=53.25°\\c=46°30^{\prime}=46.5°$

We know that

$hav(\theta)=sin^{2}(\frac{\theta}{2})$

We shall then find for the parameters a, b and c

$hav(a)=hav(68.8)=sin^{2}(\frac{68.8}{2})\\\hspace{1.cm}=sin^{2}(34.5)=0.5649^{2}=0.3192$

$hav(b)=hav(53.25)=sin^{2}(\frac{53.25}{2})\\\hspace{1.cm}=sin^{2}(26.625)=0.4481^{2}\\\hspace{1cm}=0.2008$

$hav(c)=hav(46.5)=sin^{2}(\frac{46.5}{2})\\\hspace{1.cm}=sin^{2}(23.25)=0.3947^{2}\\\hspace{1cm}=0.1558$

Also, we have

$hav(b-c)=hav(53.25-46.5)\\\hspace{1cm}=hav(6.75)=sin^{2}(\frac{6.75}{2})\\\hspace{1.cm}=sin^{2}(3.375)=0.589^{2}\\\hspace{1cm}=0.00347$

$hav(a-c)=hav(68. 8-46.5)\\\hspace{1cm}=hav(22.3)=sin^{2}(\frac{22.3}{2})\\\hspace{1.cm}=sin^{2}(11.15)=0.1934^{2}\\\hspace{1cm}=0.0374$

$hav(a-b)=hav(68.8-53.25)\\\hspace{1cm}=hav(15.55)=sin^{2}(\frac{15.55}{2})\\\hspace{1.cm}=sin^{2}(7.775)=0.1353^{2}\\\hspace{1cm}=0.0183$

The sin of the parameters a, b and c

$sin(a)=sin(68.8°)=0.9323\\sin(b)=sin(53.25°)=0.8013\\sin(c)=sin(46.5°)=0.7254$

Now

Using the Haversine formula to solve for havA, havB and havC

$havA=\frac{0.3192-0.00347}{0.8013*0.7254}=0.5432\\$

$havB=\frac{0.2008-0.00347}{0.9323*0.7254}=0.2416$

$havC=\frac{0.1558-0.0183}{0.9323*0.8013}=0.1840$

$havA=0.5432\\\implies0.5432=sin^{2}(\frac{\theta}{2})\\\implies\sqrt{0.5432}=sin(\frac{\theta}{2})\\\implies0.7370=sin(\frac{\theta}{2})\\\implies sin^{-1}(0.7370)=\frac{\theta}{2}\\\implies 47.4764=\frac{\theta}{2}\\\implies \theta=94.95° \\A=94.95°$

$havB=0.2416\\\implies0.2416=sin^{2}(\frac{\theta}{2})\\\implies\sqrt{0.2416}=sin(\frac{\theta}{2})\\\implies0.4915=sin(\frac{\theta}{2})\\\implies sin^{-1}(0.4915)=\frac{\theta}{2}\\\implies 29.44=\frac{\theta}{2}\\\implies \theta=58.88° \\B=58.88°$

$havC=0.1840\\\implies0.1840=sin^{2}(\frac{\theta}{2})\\\implies\sqrt{0.1840}=sin(\frac{\theta}{2})\\\implies0.4289=sin(\frac{\theta}{2})\\\implies sin^{-1}(0.4289)=\frac{\theta}{2}\\\implies 25.4011=\frac{\theta}{2}\\\implies \theta=50.80° \\C=50.80°$

(2) We have a right spherical triangle with

$c=69°25`, B=63°25`, A=90°$

In the first

$Sin(c)=tan(b) * tan(90-B)$

$\therefore sin(c)=tan(b) * cot(B)\\\therefore tan(b)=\frac{sin(c)}{cot(B)}=sin(c) * tan(B)$

$=sin(69.42)-tan(63.42)\\=1.8711$

$\therefore tan(b)=1.8711\\b=tan^{-1}1.8711\\b=61.88$

In the second

$sin(90-C)=cos(c) * cos(90-B)\\\therefore cos(C)= cos(c) * sin(B)\\\therefore cos(C)=cos(69.42) * sin(63.42)$

$=0.3143$

$C=cos^{-1}0.3143\\C=71.68°$

In the third

$sin(90-B)=tan(c) * tan(90-a)\\\therefore cos(B)=tan(c) * cot(a)\\\therefore cos(B)=\frac{tan(c)}{tan(a)}\\\therefore tan(a)=\frac{tan(c)}{tan(B)}$

$=\frac{tan(69.42)}{cos(63.42)}=\frac{2.6633}{0.4474}=5.9528$

$a=tan^{-1}(5.9528)\\a=80.46°$

(3) we shall calculate the Area of the spherical triangle in (1) and (2)

To calculate the Area for (1)

Using the formula

$A=\frac{\pi R^{2} E}{180}$

Where E is the excess and we have

$E= A+B+C - 180°$

To calculate the Excess for question (1)

$E=94.95°+58.88°+50.80° - 180°\\E=24.63$

Now

$A=\frac{(3.142) (30)^{2} (24.63)}{180}$

$A=\frac{(3.142) (900) (24.63)}{180}$

$A=\frac{69648.714}{180}\\A=386.9$

Also, for question (2)

$E=90°+63.42°+71.68°-180°\\E=45.1$

$A=\frac{\pi R^{2} E}{180}$

$A=\frac{(3.142) (30)^{2} (45.1)}{180}$

$A=\frac{(3.142) (900) (45.1)}{180}$

$A=\frac{127533.78}{180}\\A=708.521$