776 



Comparison of the Areas of Plane and Spherical Triangles, By Captain 

 Shortrede, 1st Assistant, Grand Trigonometrical Survey. 



[N. B. — The first part of the following investigation is taken from 

 Young's Trigonometry, but the formula there deduced 



cot ^ E= J • — — — -I- 1 ? cot C, being inconvenient in all cases, 



\_ cos C r J 



and utterly unworkable when C = 90°, I have transformed it as follows.] 



E being the spherical excess 



tan i E = tan £ (A + B -f C — 180°) == — cot £ (A + B + C) 



= tan £ C — cot ± (A + B) 



1 + taniCcot£(A+ B) 



By Napier's analogies, cot \ (A + B) = cos ' ^ + ^ 



cos \ (a — b) 

 which substituted gives 



tan|C- C0S f ra+ f taniC 

 tanlE= . cos * a ~* 



cos \ a — b 

 {cos \ (a — b) — cos \ (a -f bj) tan \ C 

 cos \ (a — b) + cos \ (a -f bj tan 2 £ C 

 cos \ (a — b) — cos \ (a -f- b) 



cos \ (a — b) cot ■£■ C + cos J fa -f- b) tan | C 

 Multiplying the numerator by sin C, and the denominator by its 

 equal 2 sin £ C cos \ C, we have 



cos \ {cos \ (a — b) — cos \ (a -f- bj) sin C 



tan \ E = 



2 cos 2 



and substituting for cos \ (a — b) and cos \ (a -f- &^, it becomes 



~ 2 cos i (a — ^ cos 2 i C + 2 cos £ fa + *>> sin 2 \ C 



, „ sin 1 a sin 4 6 sin C 



tan 1 E = 



cos \ a cos \ b + sin \ a sin £ b (cos 2 £ C — sin 2 \ C) 



and, because cos 2 \ C — sin 2 £ = cos C, 



, _ tan £ a tan 4- b sin C 



tan i E = ?-= ~-r H - 



1 -f tan ^ a tan ^ 6 cos O 



This expression has some analogy to that for the area of a plane 



triangle, but here, unlike the case of the plane triangle, it is not a 



matter of indifference whether the contained angle be acute or obtuse. 



The second term in the denominator is + or — according as C L 90 or 



C 7 90. Hence the area and excess also of a spherical triangle whose 



