The Elements of Plane TrigonometryJ. Weale, 1854 - 119 páginas |
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Página 1
... shew even in this rudimentary treatise . 2. In estimating angular measures , we suppose the right angle to be the primary one , and to be divided into 90 equal parts , each of which is called a degree ; each degree is supposed to be ...
... shew even in this rudimentary treatise . 2. In estimating angular measures , we suppose the right angle to be the primary one , and to be divided into 90 equal parts , each of which is called a degree ; each degree is supposed to be ...
Página 4
... shew that F ' = 3.32 F " : E ' 2.5 E " F ' = = 3.33 2.53 • 1 French min .; .. F ' × 100 × 100 = a quadrant , E ' = 1 English min . ; .. Ex 60 x 90 = a quadrant , hence , Fx10000 = E ' × 5400 , Fx 50 F ' - 27 E ' 50 = = E ' x 27 , 3.32 ...
... shew that F ' = 3.32 F " : E ' 2.5 E " F ' = = 3.33 2.53 • 1 French min .; .. F ' × 100 × 100 = a quadrant , E ' = 1 English min . ; .. Ex 60 x 90 = a quadrant , hence , Fx10000 = E ' × 5400 , Fx 50 F ' - 27 E ' 50 = = E ' x 27 , 3.32 ...
Página 10
... shew that sin A = sin ( 180 ° – A ) . Let BOP , be greater than a right angle ; draw PN , perpen- dicular to AB ; make the angle BOP equal to the angle AOP1 , and OP = OP1 , and let fall the A perpendicular PN , then the tri- angles PON ...
... shew that sin A = sin ( 180 ° – A ) . Let BOP , be greater than a right angle ; draw PN , perpen- dicular to AB ; make the angle BOP equal to the angle AOP1 , and OP = OP1 , and let fall the A perpendicular PN , then the tri- angles PON ...
Página 18
... Shew that sec2 A cosec3 A = sec2 A + cosec2 A. sec2 A cosec2 4 = sec3 A ( 1 + cot3 A ) = sec2 A + sec2 A cot3 A , = sec A + 1 cos2 A cos2 A sin3 A sec3 A + -- = sec2A + cosec2 A , 1 sin A Ex . 8. If tan3 A + 4 sin2 A 18 TRIGONOMETRY .
... Shew that sec2 A cosec3 A = sec2 A + cosec2 A. sec2 A cosec2 4 = sec3 A ( 1 + cot3 A ) = sec2 A + sec2 A cot3 A , = sec A + 1 cos2 A cos2 A sin3 A sec3 A + -- = sec2A + cosec2 A , 1 sin A Ex . 8. If tan3 A + 4 sin2 A 18 TRIGONOMETRY .
Página 35
... — √ / ( 1 − sin 30 ° ) } - - { { √ / ( 1 + 4 ) - { √ / ( 1-4 ) } = 1 / √3 2 12 一= √3-1 2√2 This might have been found by putting sin 150 = sin ( 45o - 30 ° ) . ( 1 ) Shew that Examples . cos ( 60 TRIGONOMETRY . 35.
... — √ / ( 1 − sin 30 ° ) } - - { { √ / ( 1 + 4 ) - { √ / ( 1-4 ) } = 1 / √3 2 12 一= √3-1 2√2 This might have been found by putting sin 150 = sin ( 45o - 30 ° ) . ( 1 ) Shew that Examples . cos ( 60 TRIGONOMETRY . 35.
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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.