Answer to Question #139004 in Trigonometry for riz

Question

Find the missing parts of an isosceles spherical triangle.
1.) 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

$1.)a = b = 78°20’’ C = 118°50’$

Cosine Law for side

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

Given a=b

Therefore

$\cos c=\cos² a+\sin² a \cos C$

$\cos c=\cos^278°20’’ +\sin^2 78°20’’\cos 118°50’\approx$$-0.4182$

$c\approx114°43'26''$

Sine law

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

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

$\sin A=\sin B= \frac{\sin 78°20’’}{\sin 114°43'26''}{\sin 118°50'}$

$A=B=70°37'35''$

Or

$A=B=109°22'25''$

$2.) A = B = 95°5' C = 100°10'$

Cosine law of angles

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

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

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

$a=\cos^—¹=\frac{\cos 95°5'+\cos 95°5' \cos 100°10'}{\sin 95°5'\sin 100°10'}$

$a\approx94°16'5''$

$c=\cos^—¹=\frac{\cos 100°10' +\cos^² 95°5'}{\sin^²95°5'}$

$c\approx99°47'15''$

$3.) B = 72°48' b = 64°52'$

$a=b = 64°52'$

Napier's analogies

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

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

$c=2\tan^—¹=(\cos (72°48')\tan(64°52'))$

$c\approx64°26'51''$

Sine law

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

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

$C=\sin^—¹(\frac{\sin 72°48' \sin 64°26'51'' }{\sin 64°52' })$

$C \approx72°10'15''$

Or

$C\approx107°49'45''$

$4.) A = C = 50°10' c = 95°$

$a=c=95°$

Napier's analogies

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

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

${\tan (\frac{1}{2}b)}= \cos 50°10'\tan 95°\approx -7.0108$

$\frac{1}{2}{b}>=90°>=b>180°$

$a+b+c>95°+95°+180°=370°$

Therefore the isosceles spherical triangle with A = C = 50°10’ c = 95° does not exist.

$5.) B = C = 78°44' b = 18°16'$

$c=b=18°16'$

Napier's analogies

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

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

$a=2\tan^—¹=(\cos (78°44')\tan(18°16'))$

$a=7°22'47''$

Sine law

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

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

$A=\sin^—¹(\frac{\sin 78°44' \sin 7°22'51'' }{\sin 18°16' })\approx23°12'1'$

$A=23°12'1'$

Or

$A=156°47'59''$