Answer to Question #138994 in Trigonometry for jay nerves

Question

Find the missing parts of an isosceles spherical triangle.
1.) 6. a = b = 78°20’’ C = 118°50’
2.) A = B = 95°5’ C = 100°10’
3.) B = 72°48’ b = 64°52’
4.) A = C = 50°10’ c = 95°
5.) B = C = 78°44’ b = 18°16’

Answer 1

1.) 6. a = b = 78°20’’ C = 118°50’

Cosine Law for sides

$\cos c=\cos a\cos b+\sin a\sin b\cos C$

$\cos c=\cos^2 a+\sin^2 a\cos C$

$\cos c=\cos^2 78\degree20''+\sin^2 78\degree20''\cos 118\degree50'\approx-0.4182$

$c\approx114\degree43'26''$

Sine Law

$\dfrac{\sin a}{\sin A}=\dfrac{\sin b}{\sin B}=\dfrac{\sin c}{\sin C}$

$\sin A=\sin B=\dfrac{\sin a\sin C}{\sin c}$

$\sin A=\sin B=\dfrac{\sin 78\degree20''\sin 118\degree50'}{\sin 114\degree43'26''}$

$A=B=70\degree37'35''$

Or

$A=B=109\degree22'25''$

2.) A = B = 95°5’ C = 100°10’

Cosine Law for angles

$\cos A=-\cos B\cos C+\sin B\sin C\cos a$

$\cos B=-\cos A\cos C+\sin A\sin C\cos b$

$\cos C=-\cos A\cos B+\sin A\sin B\cos c$

$a=\cos^{-1}\dfrac{\cos(95\degree5')+\cos(95\degree5')\cos(100\degree10')}{\sin(95\degree5')\sin(100\degree10')}$

$a\approx94\degree16'5''$

$b\approx94\degree16'5''$

$c=\cos^{-1}\dfrac{\cos(100\degree10')+\cos^2(95\degree5')}{\sin^2(95\degree5')}$

$c\approx99\degree47'15''$

3.) A=B = 72°48’ b = 64°52’

$a=b=64\degree52'$

Napier's analogies

$\dfrac{\cos (\dfrac{1}{2}(A-B))}{\cos (\dfrac{1}{2}(A+B))}=\dfrac{\tan (\dfrac{1}{2}(a+b))}{\tan (\dfrac{1}{2}c)}$

$\tan (\dfrac{1}{2}c)=\cos A\tan a$

$c=2\tan^{-1}(\cos (72\degree48')\tan (64\degree52'))\approx64\degree26'51''$

Sine Law

$\dfrac{\sin a}{\sin A}=\dfrac{\sin b}{\sin B}=\dfrac{\sin c}{\sin C}$

$\sin C=\dfrac{\sin c\sin B}{\sin b}$

$C=\sin^{-1}(\dfrac{\sin (72\degree48')\sin (64\degree26'51'')}{\sin (64\degree52')})$

$C\approx72\degree10'15''$

Or

$C\approx107\degree49'45''$

4.) A = C = 50°10’ c = 95°

$a=c=95\degree$

Napier's analogies

$\dfrac{\cos (\dfrac{1}{2}(A-C))}{\cos (\dfrac{1}{2}(A+C))}=\dfrac{\tan (\dfrac{1}{2}(a+c))}{\tan (\dfrac{1}{2}b)}$

$\tan (\dfrac{1}{2}b)=\cos A\tan a$

$\tan (\dfrac{1}{2}b)=\cos (52\degree10')\tan (95\degree)\approx-7.0108$

${1\over 2}b>90\degree=>b>180\degree$

$a+c+b>95\degree+95\degree+180\degree=370\degree$

The sum of the three sides of a spherical triangle is less than $360\degree.$ Therefore the isosceles spherical triangle with A = C = 50°10’ c = 95° does not exist.

5.) B = C = 78°44’ b = 18°16’

$c=b=18\degree16'$

Napier's analogies

$\dfrac{\cos (\dfrac{1}{2}(B-C))}{\cos (\dfrac{1}{2}(B+C))}=\dfrac{\tan (\dfrac{1}{2}(b+c))}{\tan (\dfrac{1}{2}a)}$

$\tan (\dfrac{1}{2}a)=\cos B\tan b$

$a=2\tan^{-1}(\cos (78\degree44')\tan (18\degree16'))\approx7\degree22'47''$

Sine Law

$\dfrac{\sin a}{\sin A}=\dfrac{\sin b}{\sin B}=\dfrac{\sin c}{\sin C}$

$\sin A=\dfrac{\sin a\sin B}{\sin b}$

$A=\sin^{-1}(\dfrac{\sin (78\degree44')\sin (7\degree22'51'')}{\sin (18\degree16')})$

$A\approx23\degree12'1''$

Or

$A\approx156\degree47'59''$