The Elements of Plane TrigonometryJ. Weale, 1854 - 119 páginas |
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Página 55
... determined ; but they may be more easily found from equa- tion ( 4 ) thus , put B for A ; a for b and b for a , and we have 2 - sin B ̧ √ { S ( S− a ) ( S − b ) ( S − c ) } . . . . . . ( e ) . ac - Also putting c for A , a for c ...
... determined ; but they may be more easily found from equa- tion ( 4 ) thus , put B for A ; a for b and b for a , and we have 2 - sin B ̧ √ { S ( S− a ) ( S − b ) ( S − c ) } . . . . . . ( e ) . ac - Also putting c for A , a for c ...
Página 72
... object 6 feet high , placed on the top of a tower , subtends an angle tan 015 , at a place whose horizontal distance from the foot of the tower is 100 feet , determine the tower's height . Let DAB = a , AD = x , AB 72 TRIGONOMETRY .
... object 6 feet high , placed on the top of a tower , subtends an angle tan 015 , at a place whose horizontal distance from the foot of the tower is 100 feet , determine the tower's height . Let DAB = a , AD = x , AB 72 TRIGONOMETRY .
Página 77
... determine the degree of approximation 1 1 1 5 ; sin a = 2 sina cosa , and tana > ; sin or COS 1 122 a 1 1 2 1 > a ; .. sina > a cosa ; or 2 sina > a cosa ; 1 1 and 2 sina cosa > a cosa , or sin a > a cos2 2a , or sin a > a cos2 la ; 2 1 ...
... determine the degree of approximation 1 1 1 5 ; sin a = 2 sina cosa , and tana > ; sin or COS 1 122 a 1 1 2 1 > a ; .. sina > a cosa ; or 2 sina > a cosa ; 1 1 and 2 sina cosa > a cosa , or sin a > a cos2 2a , or sin a > a cos2 la ; 2 1 ...
Página 83
... determine the sides , a + b + c ab S = = A : Sr , , 2 2 ab = 2 Sr , a + b + c = 2S , a + b = 2 S − c , a + b = 28 - c = 2S ( Sr ) = S + r , - a + b = S + r , a2 + 2ab + b2 = S2 + 2 Sr + r2 , 4ab = 8 Sr ; a2 - 2ab + b2 = S2 - 6Sr + r2 , a - ...
... determine the sides , a + b + c ab S = = A : Sr , , 2 2 ab = 2 Sr , a + b + c = 2S , a + b = 2 S − c , a + b = 28 - c = 2S ( Sr ) = S + r , - a + b = S + r , a2 + 2ab + b2 = S2 + 2 Sr + r2 , 4ab = 8 Sr ; a2 - 2ab + b2 = S2 - 6Sr + r2 , a - ...
Página 97
... determine the height of the object , and its distance from each of the three stations B , P , C. Investigation . Since AB = OA cot OBA , AP = OA cot OPA , AC - OA cot OCA , BP2 + AP2 - AB2 cos BPA = 2AP.BP and cos CAP , which is equal ...
... determine the height of the object , and its distance from each of the three stations B , P , C. Investigation . Since AB = OA cot OBA , AP = OA cot OPA , AC - OA cot OCA , BP2 + AP2 - AB2 cos BPA = 2AP.BP and cos CAP , which is equal ...
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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.