| Charles Hutton - 1811 - 424 pages
...the quotient of the cube of the length of the arc divided by the square of the radius. PROBLEM VIII. **It is required to Investigate a Theorem, by means...which, for the sake of brevity, we represent by a,** /3, y, respectively : then, by , COS a — COS & . COS y euua. I chap, iv, we have cos A n - : —... | |
| Charles Hutton - 1811 - 404 pages
...considering the Chords of the respective Arcs or Sides. bet a, b, c, be the sides, and А, в, с, **the angles of a spherical triangle, on the surface of a sphere whose radius is r ; then** then a similar triangle on the surface of a sphere whose radius = 1 , will have for its sides — ,... | |
| Charles Hutton - 1822 - 680 pages
...Plane Trigonometry, without considering the Ghofds of the respective Arcs or Sides. , 81 Let 0, 5, **c, be the sides, and A, B, c/ the angles of a spherical...the surface of a sphere whose radius is r • then a** siniilar triangle on the surface of a sphere whoseradius =3= 15 will have for its sides -, -, - ; which,... | |
| Dionysius Lardner - 1828 - 430 pages
...case, we shall establish it by geometrical construction, and subsequently derive all others from it. **Let a, b, c, be the sides, and A, B, c, the angles of a spherical triangle,** as usual. From the vertex of the angle c let tangents be drawn to the arcs a and b ; and from the centre... | |
| Euclid, Dionysius Lardner - 1828 - 542 pages
...sides, three times the sum of the squares of the sides is equal to four times that of the bisectors. **Let A, B, C, be the sides and a, b, c the** corresponding bisectors. The sum of the squares of B an'd C is equal twice the sum of the squares of... | |
| 1832 - 636 pages
...shall establish it by geometrical construction, and subsequently derive all others from it. Let a, 6, **c, be the sides, and A, B, C, the angles of a spherical triangle,** as usual. From the vortex of the angle C let tangents be drawn to the arcs a and 6; and from the centre... | |
| Cambridge Philosophical Society - 1838 - 618 pages
...example, by one which I proposed in the Transactions of the Royal Society of Edinburgh, Vol. x. viz., **Let a, b, c, be the sides, and A, B, C the** opposite angles; a+b : ab = tan^(A+B) : tan^(AB), cos±(AB) : cos^(A + B) = a + b : C ; Also sin^(AB)... | |
| Richard Abbatt - 1841 - 234 pages
...and the included side to find the other sides and the third angle. Take the polar triangle (81.) and **let a', b', c', be the sides and A' B' C' the angles** opposite : then since the sides and angles of the polar triangle are the supplements of the angles... | |
| 1856 - 410 pages
...[SECOND SOLUTION. Mr. Andrew Roy, Dundee Academy ; and similarly by Mr. Stephen Watson, Castleside.] **Let a, b, c be the sides, and A, B, C the** opposite angular points of the given triangle, and G the centre of the inscribed circle ; then if af... | |
| Royal Military Academy, Woolwich - 1853 - 476 pages
...before. or<j!> =49 8' 19" '7 2. Given A = 5T 30', C = 131 30', and b = 80 19', to find the other parts. **Let a', b', c' , be the sides, and A', B', C', the angles, of** the polar triangle, then a'= 180' -A = 128" 30', c' = 180 - C = 48' 30', B' = 180"- b = 99 41'. Whence... | |
| |