The Elements of Plane TrigonometryJ. Weale, 1854 - 119 páginas |
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Página 51
... ; for it is clear that if draw an indefinite number of lines parallel to the base AB of the triangle ABC , all the triangles so formed will have their you angles equal , but the triangles may have their sides D2 TRIGONOMETRY . F 51 CHAPTER ...
... ; for it is clear that if draw an indefinite number of lines parallel to the base AB of the triangle ABC , all the triangles so formed will have their you angles equal , but the triangles may have their sides D2 TRIGONOMETRY . F 51 CHAPTER ...
Página 52
... triangle ABC , let fall the perpendicular CD , then , by the definitions , CD CD = = sin A , and - sin B , AC CB A b a D B dividing the former by the latter , we have sin A CD CB = CB sin B AC CDAC i . e . sin A sin B :: CB : AC :: a ...
... triangle ABC , let fall the perpendicular CD , then , by the definitions , CD CD = = sin A , and - sin B , AC CB A b a D B dividing the former by the latter , we have sin A CD CB = CB sin B AC CDAC i . e . sin A sin B :: CB : AC :: a ...
Página 59
... ABC or A'BC may be the triangle required . 21. To find the area of a triangle in the terms of the radius of the inscribed circle , and sides of the triangle . The triangle ACB is composed of the triangles AOB , AOC , and BOC ; O being ...
... ABC or A'BC may be the triangle required . 21. To find the area of a triangle in the terms of the radius of the inscribed circle , and sides of the triangle . The triangle ACB is composed of the triangles AOB , AOC , and BOC ; O being ...
Página 60
James Hann. 22. To find the area of the triangle in terms of the radius of the ... abc = 2 OA 2 2 OA 4R ' We may now find the radius of the circumscribing ... triangle if a , b , c be the sides , and A , B , C the angles , prove that c ...
James Hann. 22. To find the area of the triangle in terms of the radius of the ... abc = 2 OA 2 2 OA 4R ' We may now find the radius of the circumscribing ... triangle if a , b , c be the sides , and A , B , C the angles , prove that c ...
Página 62
... triangle , and p the perpendicular from one of the equal angles upon the op- posite side : shew that the area = pc2 ... triangle ABC prove that = sin ( A - B ) a2 - b2 = 1 tan 0 , 20 / sin C sin ( A – B ) sin C sin A 62 TRIGONOMETRY .
... triangle , and p the perpendicular from one of the equal angles upon the op- posite side : shew that the area = pc2 ... triangle ABC prove that = sin ( A - B ) a2 - b2 = 1 tan 0 , 20 / sin C sin ( A – B ) sin C sin A 62 TRIGONOMETRY .
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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.