Answer to Question #139004 in Trigonometry for riz

Question #139004
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’
1
Expert's answer
2020-10-22T14:13:02-0400

"1.)a = b = 78\u00b020\u2019\u2019\tC = 118\u00b050\u2019"

Cosine Law for side



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

Given a=b

Therefore

"\\cos c=\\cos\u00b2 a+\\sin\u00b2 a \\cos C"

"\\cos c=\\cos^278\u00b020\u2019\u2019\t+\\sin^2 78\u00b020\u2019\u2019\\cos 118\u00b050\u2019\\approx""-0.4182"

"c\\approx114\u00b043'26''"

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\u00b020\u2019\u2019}{\\sin 114\u00b043'26''}{\\sin 118\u00b050'}"

"A=B=70\u00b037'35''"


Or

"A=B=109\u00b022'25''"



"2.) A = B = 95\u00b05'\tC = 100\u00b010'"


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^\u2014\u00b9=\\frac{\\cos 95\u00b05'+\\cos 95\u00b05' \\cos 100\u00b010'}{\\sin 95\u00b05'\\sin 100\u00b010'}"


"a\\approx94\u00b016'5''"

"c=\\cos^\u2014\u00b9=\\frac{\\cos 100\u00b010' +\\cos^\u00b2 95\u00b05'}{\\sin^\u00b295\u00b05'}"


"c\\approx99\u00b047'15''"


"3.) B = 72\u00b048'\tb = 64\u00b052'"

"a=b = 64\u00b052'"


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^\u2014\u00b9=(\\cos (72\u00b048')\\tan(64\u00b052'))"

"c\\approx64\u00b026'51''"

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^\u2014\u00b9(\\frac{\\sin 72\u00b048' \\sin 64\u00b026'51'' }{\\sin 64\u00b052' })"

"C \\approx72\u00b010'15''"

Or


"C\\approx107\u00b049'45''"


"4.) A = C = 50\u00b010'\tc = 95\u00b0"

"a=c=95\u00b0"

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\u00b010'\\tan 95\u00b0\\approx -7.0108"


"\\frac{1}{2}{b}>=90\u00b0>=b>180\u00b0"


"a+b+c>95\u00b0+95\u00b0+180\u00b0=370\u00b0"


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

"5.) B = C = 78\u00b044' b = 18\u00b016'"

"c=b=18\u00b016'"

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^\u2014\u00b9=(\\cos (78\u00b044')\\tan(18\u00b016'))"


"a=7\u00b022'47''"

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^\u2014\u00b9(\\frac{\\sin 78\u00b044' \\sin 7\u00b022'51'' }{\\sin 18\u00b016' })\\approx23\u00b012'1'"


"A=23\u00b012'1'"

Or

"A=156\u00b047'59''"



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