The Elements of Plane TrigonometryJ. Weale, 1854 - 119 páginas |
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Página 29
... gives the greater , and half the difference subtracted from half the sum gives the less , we have A = ↓ ( A + B ) + † ( A − B ) , - - B = 1⁄2 ( A + B ) − 1 ( A – B ) . Putting ( A + B ) for A , and - ( A – B ) for B , in sin ( A + B ) ...
... gives the greater , and half the difference subtracted from half the sum gives the less , we have A = ↓ ( A + B ) + † ( A − B ) , - - B = 1⁄2 ( A + B ) − 1 ( A – B ) . Putting ( A + B ) for A , and - ( A – B ) for B , in sin ( A + B ) ...
Página 53
... gives AC2 = AB2 + BC2 - 2AB . BC cos B , or b2 = c2 + a23 - 2ac . cos B ; a2 + c2 - b2 ... cos B = 2ac Now , by putting c for a , and a for c , and A for B , we have b3 + c2 - a3 cos A 2bc cos C a2 + b2 → c2 - = 2ab In the same manner ...
... gives AC2 = AB2 + BC2 - 2AB . BC cos B , or b2 = c2 + a23 - 2ac . cos B ; a2 + c2 - b2 ... cos B = 2ac Now , by putting c for a , and a for c , and A for B , we have b3 + c2 - a3 cos A 2bc cos C a2 + b2 → c2 - = 2ab In the same manner ...
Página 56
... gives Δ △ = 1 / AB.AC . 1 2 - AB.AC.Z √ { S ( S − a ) ( S —b ) ( S − c ) } 2 = = bc . Z √ ↓ { S ( S − a ) ( S − b ) ( S – c ) } 2 bc - - – = √ { S ( S − a ) ( - S − b ) ( S — c ) } . = - - 19. When two sides and their ...
... gives Δ △ = 1 / AB.AC . 1 2 - AB.AC.Z √ { S ( S − a ) ( S —b ) ( S − c ) } 2 = = bc . Z √ ↓ { S ( S − a ) ( S − b ) ( S – c ) } 2 bc - - – = √ { S ( S − a ) ( - S − b ) ( S — c ) } . = - - 19. When two sides and their ...
Página 95
... gives the greater segment , and half their difference subtracted from half their sum gives the least . The following logarithmic forms for right - angled tri- angles may also be useful , although they are only the defi- nitions at page ...
... gives the greater segment , and half their difference subtracted from half their sum gives the least . The following logarithmic forms for right - angled tri- angles may also be useful , although they are only the defi- nitions at page ...
Página 97
... gives OA = = CP cot2 OBA + BP cot2 OCA - ( CP + BP ) cot2 OPA √CP co ( CP + BP ) .CP.BP E ( 3 ) If the height of the mountain called the Peak of Teneriffe be 23 miles , as it is very nearly , and the angle taken at the top of it , as ...
... gives OA = = CP cot2 OBA + BP cot2 OCA - ( CP + BP ) cot2 OPA √CP co ( CP + BP ) .CP.BP E ( 3 ) If the height of the mountain called the Peak of Teneriffe be 23 miles , as it is very nearly , and the angle taken at the top of it , as ...
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Términos y frases comunes
A+ cot a+b+c AB² AC² angle ACB angle of elevation AO² AP AP Asin BOP₁ centre circumference circumscribed cos A cos cos² cos³ cosec cosine cot A cot cot A+ cot² OPA cot³ diameter distance Divide equation feet formula given height hence hexagon inscribed circle logarithms negative nth root number of sides Oa cot OA² OB² perimeter perpendicular plane triangle quadrant r² cot radii radius ratio regular polygon right angle right-angled triangle sec² sector shew sin A cos sin A sin sin-¹ sin² sin² 18 sin³ sine sine and cosine square subtend subtracted tan-¹ tan² tan³ tangent triangle ABC unity ηφ
Pasajes populares
Página 97 - From a window near the bottom of a house, which seemed to be on a level with the bottom of a steeple, I took the angle of elevation of the top of the steeple, equal to 40° ; then from another window, 18 feet directly above the former, the like angle was 37° 30'.
Página 97 - Wanting to know the height of an inaccessible tower; at the least distance from it, on the same horizontal plane, I took its angle of elevation equal to 58°; then going 300 feet directly from it, found the angle there to be only 32°: required its height, and my distance from it at the first station?
Página 86 - The area of a regular inscribed hexagon is a mean proportional between the areas of the inscribed and circumscribed equilateral triangles.
Página 51 - It depends on the principle, that the difference of the squares of two quantities is equal to the product of the sum and difference of the quantities.
Página 63 - If from one of the angles of a rectangle a perpendicular be drawn to its diagonal, and from, the point of their intersection lines be drawn perpendicular to the sides which contain the opposite angle...
Página 87 - The square on the side of a regular pentagon inscribed in a circle is equal to the sum of the squares on the sides of the regular hexagon and decagon inscribed in the same circle.
Página 93 - To THEIR DIFFERENCE ; - So IS THE TANGENT OF HALF THE SUM OF THE OPPOSITE ANGLES; To THE TANGENT OF HALF THEIR DIFFERENCE.
Página 98 - Wanting to know my distance from an inaccessible object 0, on the other side of a river; and having no instrument for taking angles, but only a chain or cord for measuring distances; from each of two stations, A and B, which were taken at 500 yards asunder, I measured in a direct line from the object 0 100 yards, viz. AC and BD each equal to 100 yards ; also the diagonal AD measured 550 yards, and the diagonal BC 560. What then was the distance of the object 0 from each, station A and B ? . C AO...
Página 97 - Being on a horizontal plane, and wanting to know the height of a tower placed on the top of an inaccessible hill : I took the angle of elevation of the top of the hill 40°, and of the top of the tower 51°; then measuring in a line directly from it to the distance of 200 feet farther, I found the angle at the top of the tower to be 33° 45'.
Página 101 - The hypotenuse AB of a right-angled triangle ABC is trisected in the points D, E; prove that if CD, CE be joined, the sum of the squares on the sides of the triangle CDE is equal to two-thirds of the square on AB.