We will use Napier's formulas for a spherical right-angled triangle.
(More information: https://en.wikipedia.org/wiki/Spherical_trigonometry)
Hint : Since it is not specified which side is 90∘ , I chose it myself.
In our case,
1) c=90∘a=49∘23′b=76∘41′
Then,
(Q8)cosa=sinb⋅cosA→cosA=sinbcosa=sin(76∘41′)cos(49∘23′)cosA≈0.669→A=arccos(0.669)≈41∘59′24′′(Q5)tanB=tanb⋅sinA=tan(76∘41′)⋅sin(41∘59′24′′)tanB≈2.826→B=arctan(2.826)≈70∘30′40′′(Q10)cosC=−cota⋅cotb=−cot(76∘41′)⋅cot(49∘23′)cosC≈−0.203→C=arccos(−0.203)≈101∘42′
2) c=90∘B=100∘b=50∘10′
Then,
(Q9)cosb=sina⋅cosB→sina=cosBcosb=cos(100∘)cos(50∘10′)sina≈0.7428→a=arcsin(0.7428)≈47∘58′12′′(Q4)tanA=tana⋅sinB=tan(47∘58′12′′)⋅sin(100∘)tanA≈1.093→A=arctan(1.093)≈47∘32′24′′(Q1)cosC=−cosA⋅cosB→cosC=−cos(47∘32′24′′)⋅cos(100∘)cosC≈0.1172→C=arccos(0.1172)≈83∘16′12′′
3) c=90∘A=121∘20′B=42∘01′
Then,
(Q1)cosC=−cosA⋅cosB=−cos(121∘20′)⋅cos(42∘01′)cosC≈0.3863→C=arccos(0.38630)≈67∘16′50′′(Q4)tanA=tana⋅sinB→tana=sinBtanA=sin(42∘01′)tan(121∘20′)tana≈−2.454→a=arctan(−2.454)≈112∘10′10′′(Q5)tanB=tanb⋅sinA→tana=sinAtanB=sin(121∘20′)tan(42∘01′)tanb≈1.055→b=arctan(1.055)≈46∘31′50′′ 4) c=90∘a=60∘35′B=122∘18′
Then,
(Q4)tanA=tana⋅sinB→tanA=tan(60∘35′)⋅sin(122∘18′)tanA≈1.499→a=arctan(1.499)≈56∘17′24′′(Q9)cosb=sina⋅cosB→cosb=sin(60∘35′)⋅cos(122∘18′)cosb≈−0.4655→b=arccos(−0.4655)≈117∘42′(Q6)tanB=−cosa⋅tanC→tanC=−cosatanBtanC≈3.221→C=arctan(3.221)≈72∘45′
Q.E.D.
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