Elements of Geometry, and Plane and Spherical Trigonometry: With Numerous Practical ProblemsIvison, Blakeman, Taylor, 1860 - 453 sider |
Inni boken
Resultat 1-5 av 55
Side 246
With Numerous Practical Problems Horatio Nelson Robinson. 9. The Versed Sine of an ... cos . AB ; for tangent AB , we write tan . AB , etc. From the preceding ... a quadrant are each equal to the radius ; its cosine is zero , and its ...
With Numerous Practical Problems Horatio Nelson Robinson. 9. The Versed Sine of an ... cos . AB ; for tangent AB , we write tan . AB , etc. From the preceding ... a quadrant are each equal to the radius ; its cosine is zero , and its ...
Side 248
... the first , second , third , and fourth quadrants . The center of the circle is taken as ... Cos . tan . cot . sec . cosec . vers . + + + + + + + + + 4th 66 PROPOSITION ... a center , and CA as a adius 248 PLANE TRIGONOMETRY . Propositions.
... the first , second , third , and fourth quadrants . The center of the circle is taken as ... Cos . tan . cot . sec . cosec . vers . + + + + + + + + + 4th 66 PROPOSITION ... a center , and CA as a adius 248 PLANE TRIGONOMETRY . Propositions.
Side 250
... the perpendiculars IH on FM , and EK on IN . Now , by the definition of sines and cosines , DO = sin.a ; GO cos.a ; FI = sin.b ; GI cos.b. We are to find = = = FM = sin . ( a + b ) ; GM = cos . ( a + b ) ; EP = sin . ( ab ) ; GP cos ...
... the perpendiculars IH on FM , and EK on IN . Now , by the definition of sines and cosines , DO = sin.a ; GO cos.a ; FI = sin.b ; GI cos.b. We are to find = = = FM = sin . ( a + b ) ; GM = cos . ( a + b ) ; EP = sin . ( ab ) ; GP cos ...
Side 251
... a cos.b + cos.a sin.b Ꭱ By subtracting the second from the first , since IN — FH = IN- IK = EP , we have sin . ( a - b ) = - sin.a cos.b cos.a sin.b R By subtracting the fourth from the third , we have GN - IH GM = cos . ( a + b ) for ...
... a cos.b + cos.a sin.b Ꭱ By subtracting the second from the first , since IN — FH = IN- IK = EP , we have sin . ( a - b ) = - sin.a cos.b cos.a sin.b R By subtracting the fourth from the third , we have GN - IH GM = cos . ( a + b ) for ...
Side 252
With Numerous Practical Problems Horatio Nelson Robinson. A — B sin.A + sin.B = 2sin . ( 4+ B ) cos . ( AB ) ( 15 ) 2 Ᏼ 2 s : n.A —sin.B = 2cos . ( 4+ B ) sin . ( B ) ( 16 ) 2 ( 0 ) cos . A + cos . B : 2cos . = ( A + B ) cos . ( 4 — B ) ( ...
With Numerous Practical Problems Horatio Nelson Robinson. A — B sin.A + sin.B = 2sin . ( 4+ B ) cos . ( AB ) ( 15 ) 2 Ᏼ 2 s : n.A —sin.B = 2cos . ( 4+ B ) sin . ( B ) ( 16 ) 2 ( 0 ) cos . A + cos . B : 2cos . = ( A + B ) cos . ( 4 — B ) ( ...
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Elements of Geometry and Plane and Spherical Trigonometry: With Numerous ... Horatio Nelson Robinson Uten tilgangsbegrensning - 1867 |
Elements of Geometry, and Plane and Spherical Trigonometry: With Numerous ... Horatio Nelson Robinson Uten tilgangsbegrensning - 1865 |
Elements of Geometry, and Plane and Spherical Trigonometry: With Numerous ... Horatio Nelson Robinson Uten tilgangsbegrensning - 1860 |
Vanlige uttrykk og setninger
ABCD altitude angle opposite axis bisected chord circle circumference circumscribed common cone convex surface cos.a cos.b cos.c Cosine Cotang diagonal diameter dicular difference distance divided draw equal angles equation equiangular equivalent find the angles formulæ four magnitudes frustum given line greater half Hence the theorem homologous hypotenuse included angle inscribed intersect isosceles less Let ABC logarithm measured multiplied N.sine number of sides opposite angles parallelogram parallelopipedon pendicular perpen perpendicular plane ST polyedron PROBLEM produced Prop proportion PROPOSITION prove pyramid quadrantal radii radius rectangle regular polygon right angles right-angled spherical triangle right-angled triangle SCHOLIUM secant segment similar sin.a sin.b sin.c sine solid angles sphere SPHERICAL TRIGONOMETRY straight line subtracting Tang tangent three angles three sides triangle ABC triangular prisms TRIGONOMETRY vertex vertical angle volume
Populære avsnitt
Side 320 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.
Side 65 - If four magnitudes are in proportion, the sum of the first and second is to their difference as the sum of the third and fourth is to their difference.
Side 121 - In a given circle to inscribe a triangle equiangular to a given triangle. Let ABC be the given circle, and DEF the given triangle ; it is required to inscribe in the circle ABC a triangle equiangular to the triangle DEF.
Side 56 - If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section, is equal to the square of half the line.
Side 34 - Conversely: if two angles of a triangle are equal, the sides opposite to them are equal, and the triangle it itosceles.
Side 126 - To inscribe a regular polygon of a certain number of sides in a given circle, we have only to divide the circumference into as many equal parts as the polygon has sides : for the arcs being equal, the chords AB, BC, CD, &c.
Side 22 - If two parallel lines are cut by a third straight line, the sum of the two interior angles on the same side of the secant line is equal to two right angles.
Side 277 - The logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent of the power.
Side 94 - The angle in a semicircle is a right angle ; the angle in a segment greater than a semicircle is less than a right angle ; and the angle in a segment less than a semicircle is greater than a right angle.
Side 30 - Therefore all the interior angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has sides.