. Elements of plane and spherical trigonometry . have 1 . „ 2 sin hs sin h (s — a~) sin h (? — V) sin h (s — c) I — sm £= — . cos ha cos ho cos he 168 SPHERICAL TRIGONOMETRY. Substituting this value in the numerator and the value of —cos S from (160) in the denominator of the expression for tanIK, we have, after reduction, b) sin j (s — c)n tan ! K 4 sin Is sin | (s — a) sin \ {t n _ 4 sin Is sin \ (s — a) sin \ (s — b) sin j (s — c) Now, l/sin s sin (s — a) sin (s — 6) sin (s — c)sin 55 sin is V sin s y 2 sin Js cos %s —77, l/tan £s, .-. tan £if=i/ tan Jstan^ (s—a)tan£ (s — b) tan f (
. Elements of plane and spherical trigonometry . have 1 . „ 2 sin hs sin h (s — a~) sin h (? — V) sin h (s — c) I — sm £= — . cos ha cos ho cos he 168 SPHERICAL TRIGONOMETRY. Substituting this value in the numerator and the value of —cos S from (160) in the denominator of the expression for tanIK, we have, after reduction, b) sin j (s — c)n tan ! K 4 sin Is sin | (s — a) sin \ {t n _ 4 sin Is sin \ (s — a) sin \ (s — b) sin j (s — c) Now, l/sin s sin (s — a) sin (s — 6) sin (s — c)sin 55 sin is V sin s y 2 sin Js cos %s —77, l/tan £s, .-. tan £if=i/ tan Jstan^ (s—a)tan£ (s — b) tan f (s— c), (162)which is known as Lhuilliers formula. 125. Circumscribed Circle. Let 0 be the pole of the smallcircle circumscribed about the tri-angle ABC Draw the arcs OA,OB, and OC, and draw OP from0 perpendicular to BG. The tri-angles OBC, OCA, and OAB areisosceles, and BP=%a. Let Itrepresent the radius OA, OB, orOC In the right triangle OBPwe have, by § 107, II., cos OBP = cot B tan \a,tan \a Fig. or Now, and Adding, tan .B = cos OBP OBP = B — ABO=B — BAO, OBP= OCP= C—ACO= C — OAC. 20BP = B-\- C—(BAO+ OAC), = B+ C—A, = 2(S — A): OBP=S—A. AREA, CIRCUMSCRIBED AND INSCRIBED CIRCLES. 169 Hence tan R = tan la tan R = tan R cos (S—Ay tan lb cos (S—B)tan %c (163) cos (£— C) If the value of tan \a, from (139), be substituted in the first ofthese expressions, we have1 tan R = cos (S—A) j or tan R -v cos S cos (ff— A)cos (S—B) cos (£— C) (164) cos £ 0) Fig. 42. cos (S—A) cos (S—B) cos (&? 126. Inscribed Circle. Let 0 be the pole of the circle inscribed in the triangleABC. Draw the arcs ON, OM, and OQ perpendicular toAB, BG, and CA, at the points where theyare tangent to the inscribed circle; anddraw OA, OB, OC. The right trianglesAON and AOQ have their correspondingparts mutually equal, since they have theside A0 in common and NO equal to AN= A Q. Similarly BN= BM,and CM = CQ. We have also OA,
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