The Elements of Plane TrigonometryJ. Weale, 1854 - 119 páginas |
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Página
... sine , cosine , tangent , cotangent , secant and cosecant of 30 ° , 45 ° , and 60o CHAPTER II . Page 1 2 , 3 • 8 12 , 13 · 14 , 15 sin ( A + B ) = sin A cos B ± cos A sin B cos ( A + B ) = cos A cos B sin A sin B sin 3 A , cos 3A , & c ...
... sine , cosine , tangent , cotangent , secant and cosecant of 30 ° , 45 ° , and 60o CHAPTER II . Page 1 2 , 3 • 8 12 , 13 · 14 , 15 sin ( A + B ) = sin A cos B ± cos A sin B cos ( A + B ) = cos A cos B sin A sin B sin 3 A , cos 3A , & c ...
Página 8
... sine of A ; is the cosine of A ; A DP AP AD AD DP AD is the tangent of A ; is the cosecant of A ; AP DP AD is the secant of A ; AP AP DP is the cotangent of A. D B P C For the sake of abbreviation the above quantities are generally put ...
... sine of A ; is the cosine of A ; A DP AP AD AD DP AD is the tangent of A ; is the cosecant of A ; AP DP AD is the secant of A ; AP AP DP is the cotangent of A. D B P C For the sake of abbreviation the above quantities are generally put ...
Página 9
... sine of A , by the AD AD above definitions ; cos2 A + sin3 A = 1 ........ . ( 2 ) . From this equation we have , cos2 A = 1- sin2 A ; ..cos A = √ ( 1- sin3 A ) , and sin3 A = 1- cos2 A , sin A ( 1 - cos3 A ) ; DP and since by the ...
... sine of A , by the AD AD above definitions ; cos2 A + sin3 A = 1 ........ . ( 2 ) . From this equation we have , cos2 A = 1- sin2 A ; ..cos A = √ ( 1- sin3 A ) , and sin3 A = 1- cos2 A , sin A ( 1 - cos3 A ) ; DP and since by the ...
Página 10
... sine of an angle is equal to the cosine of its complement . Since ADP = 90o — A , - AP and sin ADP : = = cos A , AD DP cos ADP = = sin A ; AD that is , sin ( 90 ° – A ) = cos A , - and cos ( 90 ° - A ) = sin A. - ( 2 ) To shew that sin ...
... sine of an angle is equal to the cosine of its complement . Since ADP = 90o — A , - AP and sin ADP : = = cos A , AD DP cos ADP = = sin A ; AD that is , sin ( 90 ° – A ) = cos A , - and cos ( 90 ° - A ) = sin A. - ( 2 ) To shew that sin ...
Página 11
... sine of an angle is equal to the sine of the sup- plement of that angle : cos BOP1 = ON , ON = OP OP ' ON cos BOP = cos ( 180 ° -BOP ) = OP ..cos BOP1 = cos ( 180 ° - BOP , ) ; or the cosine of an angle is equal to minus the cosine of ...
... sine of an angle is equal to the sine of the sup- plement of that angle : cos BOP1 = ON , ON = OP OP ' ON cos BOP = cos ( 180 ° -BOP ) = OP ..cos BOP1 = cos ( 180 ° - BOP , ) ; or the cosine of an angle is equal to minus the cosine of ...
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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.