This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1889 Excerpt: ...incircle of a spherical triangle ABC. D Fig. 27. Sol.--Bisect the angles A, 2? by the arcs AO, BO. 0 is the incentre required; and the perpendiculars OB, OF, OF on the sides are the angular radii. Dem.--It is easy to see that OB, OF, OF are all equal; also that AF= s-a. Now from the right-angled triangle OAF-wehave, ...
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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1889 Excerpt: ...incircle of a spherical triangle ABC. D Fig. 27. Sol.--Bisect the angles A, 2? by the arcs AO, BO. 0 is the incentre required; and the perpendiculars OB, OF, OF on the sides are the angular radii. Dem.--It is easy to see that OB, OF, OF are all equal; also that AF= s-a. Now from the right-angled triangle OAF-wehave, equation (110), tan OAF= tan OFsin AF; or, denoting the radius by r, tan iA = tan r-sin (s-a). (299) Hence, substituting for tan A its value, equation (22), we get tanr= fgingfj-) sin (--c) _ _ (g00) sin- sin s Cor. 1.--If in the expression for tan r we substitute for 6, c (it-b), (ir-c), we get the expression for the in-radius of the colunar triangle BCA' formed by producing the sides AB, AC.. Hence, denoting it by r, we get tan ra =-. (301) sin (--a) Similarly, tan rb = sin_- (302) and tan r. =--, --r. (303) sin (--c) Cor. 2.--From the equations (299), (300) we get the following formulae for solving a spherical triangle when the three sides are given: --I tan r = I sin (s-a) + I sin (s-b) + I sin (--c)-I sins). (304) I tan l, A = I tan r-I sin (--a), &c. (305) Def. XXV.--The incircles of the colunar triangles are called escribed circles. 15. Prove that the cosines of the angles of the triangle Oa, Ob, Oe are i espectively equal to cos -. sin J.4, cos. sin J B, cos s. sin C. Section III.--Circumctrcles. 75. To find the circumradius of a spherical triangle ABC. Sol.--Bisect the arcs BC, CA sXB, E, and let 0 be the intersection of perpendiculars to BC, CA, at B, E; then 0 is the circumcentre. Cor. l.--If in equation (319) we substitute for B, C, Tt-B, It-C, respectively, we get the eircumradius of the colunar triangle BCA'. Hence, denoting it by RA, we have N 'sin A-E)' 10. Prove tan ...
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Add this copy of A Treatise on Spherical Trigonometry and Its to cart. $16.27, new condition, Sold by Ingram Customer Returns Center rated 5.0 out of 5 stars, ships from NV, USA, published 2022 by Legare Street Press.
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