LEGENDRE THEOREM ON SPHERICAL TRIANGLES

Legendre theorem on spherical triangles50 years of the Research Institute ofGeodesy, Topography and CartographyIn the same way one finds that the second 3 in (Ga) has the form 1 + (4) . – As [remember that the2excess is ε = (4) ]then22 * ⎡ ∗ ε ∗ ⎤ 2 ∗ ε ∗ ∗sin ( α − ε / 6) = ⎢sinα − cos α + (4) = sin α − sinα cos α + (4) ,6 ⎥⎣⎦3* * ε *sin( α + ε / 3) = sinα + cos α + (4) ,32 ∗∗3 *sin ( α − ε / 6) sin( α + ε / 3) sin α + (4)= = 1 + 43 * 3 *sin αsin αand the third 3 in (Ga) has the form 1 + (4) again. Quite analogously the same holds for the fourth3in (Ga). – The product of all roots in (Ga) thus has the form 1 + (4) and Gauss’ expression of theLegendre theorem (L) may be expressed as follows:a sin( α − ε / 3)= ⋅ {1 − (4)}b sin( β − ε / 3)or – using the cyclic changea: b: c= sin( α −ε /3):sin( β −ε /3):sin( γ −ε/3) .If a= b= c, then of course α = β = γ (and vice versa) and in Gauss’ formula (Ga) in this casethe product of the four cube roots equals 1. Problem I is closely connected withProblem II: Determine all spherical triangles for which the product of the four cube roots in theformula (Ga) equals exactly or approximately 1 .III.Let ∆ be the area of plane triangle ABC ′ ′ ′ with sides a , b , c ; let p be a parameter in theneighbourhood of null. In [9] it was proved that⎡ ⎛1⎞ ⎤sin ⎢α− ⎜ + 2 p ⎟ε3⎥23a+ pa + (5) abc⎣ ⎝ ⎠ ⎦ ∆2 2 2(N) = { 1 − p( a + b + c ) + (4)}The fraction on the right-hand side equals the reciprocal diameter 2r of the circle circumscribed abouttriangle ABC ′ ′ ′; let us remind the reader thatsinα′ 1 2∆= = .a 2r abcAs the right-hand side in (N) does not change – with the exception of (4) – with the cyclic change ofsides a , b , c , then – again with the exception of (4) –( ).( N’ )⎡ ⎛1 ⎞ ⎤ ⎡ ⎛1 ⎞ ⎤ ⎡ ⎛1⎞ ⎤sin ⎢α − ⎜ + 2 p⎟ε sin β 2 p ε sin γ 2 p ε3⎥ ⎢ − ⎜ + ⎟3⎥ ⎢ − ⎜ + ⎟3⎥⎣ ⎝ ⎠ ⎦ ⎝ ⎠ ⎝ ⎠=⎣ ⎦ ⎣ ⎦ = ;3 3 3a+ pa + (5) b+ pb + (5) c+ pc + (5)the meaning of the symbol will not be repeated.45