. A treatise on plane and spherical trigonometry, and its applications to astronomy and geodesy, with numerous examples . sines. — In any triangle the square of anyside is equal to the sum of the squares of the other two sidesminus twice the product of these sides and the cosine of theincluded angle. In an acute-angled triangle (seefirst figure) we have (Geom., BookIII., Prop. 26) BC2 = AC2 + AB2 - 2 AB x AD,or a2 = 62 + AD = b cos A. ,\ a2 = b2 + c2 — 2 be cos A. In an obtuse-angled triangle (seesecond figure) we have (Geom.,Book III., Prop. 27) BC2 = AC2 + AB2 + 2 AB x AD, or a2
. A treatise on plane and spherical trigonometry, and its applications to astronomy and geodesy, with numerous examples . sines. — In any triangle the square of anyside is equal to the sum of the squares of the other two sidesminus twice the product of these sides and the cosine of theincluded angle. In an acute-angled triangle (seefirst figure) we have (Geom., BookIII., Prop. 26) BC2 = AC2 + AB2 - 2 AB x AD,or a2 = 62 + AD = b cos A. ,\ a2 = b2 + c2 — 2 be cos A. In an obtuse-angled triangle (seesecond figure) we have (Geom.,Book III., Prop. 27) BC2 = AC2 + AB2 + 2 AB x AD, or a2 = 62 + c2 + But AD = b cos CAD = — b cos A. ,\ a2= 62 + c2 — 2 6c cos A. Similarly, b2 = c2 + a2 — 2 ca cos B, c2 = a2 + &2-2a&cosC. Note. — When one equation in the solution of triangles has been obtained, theother two may generally be obtained by advancing the letters so that a becomes b,b becomes c, and c becomes a, the order is abc, bca, cab. It is obvious that theformulae thus obtained are true, since the naming of the sides makes no difference,provided the right order is OBLIQUE TRIANGLES. 149 97. Law of Tangents. — In any triangle the sum of any two sides is to their difference as the tangent of half the sumof the opposite angles is to the tangent of half their difference. By Art. 95, a:b = sin A : sin B. By composition and division, a + b _ sin A + sin Ba —? b sin A — sin B = tan*(A + B) b (13) f A 61 (1) Similarly. M^ = tanj(B+ C) (2) b-c tan£(B-C) V ; c + a _ tan j-(C + A) (~\ c-a~tan|(C — A) ^ Since tan£(A + B) = tan(90° - £C) = cot ^C, the result in (1) may be written a + b _ cot^C /n a-&tan£(A-B) ^ ) and similar expressions for (2) and (3). 98. To show that in any triangle c = a cos B + b cos A. In an acute-angled triangle (first figure of Art. 96) wenave c = DB + DA = a cos B + b cos A. In an obtuse-angled triangle (second figure of Art. 96)we have C = DB-DA = a cos B — b cos c = acosB + 6 cos , b
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Keywords: ., bookcentury1900, bookdecade1900, booksubjecttrigono, bookyear1902
