Answer to Question #138994 in Trigonometry for jay nerves

Question #138994
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’
1
Expert's answer
2020-10-21T14:20:04-0400

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''"


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