Navigation New ModelledW. and J. Mount, 1761 - 528 sider |
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Side 41
... Subtraction to be 6od . am . Then by Axiom the fecond , J d . m , As the Sine of B 45 0 C. Ar , 0.15051 To Side oppofite A C 410 2.61278 So Sine of A 60 2 0.03767 To Side oppofite 502 2.70097 But if the Angle at C had been fuppofed ...
... Subtraction to be 6od . am . Then by Axiom the fecond , J d . m , As the Sine of B 45 0 C. Ar , 0.15051 To Side oppofite A C 410 2.61278 So Sine of A 60 2 0.03767 To Side oppofite 502 2.70097 But if the Angle at C had been fuppofed ...
Side 76
... Subtract 1 Subtract the given Latitude and Middle Latitude the less from 76 Chap . II . Middle Latitude Sailing .
... Subtract 1 Subtract the given Latitude and Middle Latitude the less from 76 Chap . II . Middle Latitude Sailing .
Side 77
Henry Wilson. Subtract the given Latitude and Middle Latitude the less from the greater , the Remainder doubled is the Difference of Latitude , Then for the Course . As Difference of Latitude to Radius , fo Departure to Tangent of the ...
Henry Wilson. Subtract the given Latitude and Middle Latitude the less from the greater , the Remainder doubled is the Difference of Latitude , Then for the Course . As Difference of Latitude to Radius , fo Departure to Tangent of the ...
Side 97
... Subtract the Square of the Departure from the Square of the Distance , the Square Root of the Remainder is the Dif- ference of Latitude ; and then find the Course as in Cafe IV ; only mind to find the Quotient in the Departure Column ...
... Subtract the Square of the Departure from the Square of the Distance , the Square Root of the Remainder is the Dif- ference of Latitude ; and then find the Course as in Cafe IV ; only mind to find the Quotient in the Departure Column ...
Side 133
... Subtraction ; fo here it it performed by taking in your Compaffes the Distance upon the Scale , be- tween the first and fecond given Terms , and the fame Ex- tent will reach from the third given Term to the fourth Term required ; but ...
... Subtraction ; fo here it it performed by taking in your Compaffes the Distance upon the Scale , be- tween the first and fecond given Terms , and the fame Ex- tent will reach from the third given Term to the fourth Term required ; but ...
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Navigation New Modelled: Or a Treatise of Geometrical, Trigonometrical ... Wilson Uten tilgangsbegrensning - 1752 |
Navigation New Modelled: Or, a Treatise of Geometrical, Trigonometrical ... Henry Wilson Ingen forhåndsvisning tilgjengelig - 2018 |
Vanlige uttrykk og setninger
againſt alfo anſwer Bafe becauſe Cape Chord Column Compaffes confequently Courfe and Difference Courſe and Diſtance Cyphers Degrees demand the Courfe Dep Lat Dep Departure given Diff Difference of Latitude Difference of Lon Difference of Longitude Dift draw the Line Eaft faid fame fecond fhall find the Courfe firft firſt fo Sine fome ftance fubtract fuppofe given Number half Hypotenufe Interfection laft Lat Dep Lat Lati Latitude 50 Latitude and Departure leffer Longitude given meaſured Mercator's Sailing Merid Meridian Meridional Difference Middle Latitude Sailing Miles Minutes Sine muſt North obferve Parallel Parallel Sailing parture Perpendicular Plane Sailing Arithmetical Plane Triangles Point proper Difference Proportion Queftion Quotient Radius reft reprefent Right-angled RULE Secant Minutes Ship fails Ship in Latitude Side AC Side oppofite Sine Comp Sine Complement Sine Tangent South Square Sun's Table Tangent Secant thefe theſe tude uſe Weft ΙΟ
Populære avsnitt
Side xiv - In any plane triangle, the sum of tfte two sides containing either angle, is to their difference, as the tangent of half the sum of the other two angles, to the tangent of half their difference.
Side 185 - Multiply the moon's age by 4, and divide the product by 5 ; the quotient...
Side xiv - Then as the sum of the sides is to their difference, so is the tangent of half the sum of the unknown angles to the tangent of half their difference (Theor.
Side xiv - ... the sum of the segments of the base is to the sum of the sides as the difference of the sides to the difference of the segments of the base.
Side 231 - ... of the opposite parts. All of the ten formulas just listed can easily be derived from these rules. Oblique Spherical Triangles.— Formulas for the solution of oblique spherical triangles are developed from those for right spherical triangles. The law of sines states that in any spherical triangle the sines of the sides are proportional to the sines of their opposite angles: sin a sin b sin с...
Side xi - Circumference of every Circle is fuppofed to be Divided into 360 equal Parts called Degrees, and each Degree...
Side 24 - Then radius : log. 179-3 :: s. 8° 31' : log. ofi6'5i; which doubled is the diftance between Edyftone and Lizard 53-04 miles. For the other two, there are 2 fides, and an angle oppofite to one of them, given, to find the other fide.
Side 232 - Day, fubtradt the Complement of the Sun's Altitude, noting the half Sum, and the Remainder. 3. Then to find the Sun's Azimuth, it's thus ; to the Complement Arithmetic of the Sines of the Complement of the Latitude, and Complement of the Sun's Altitude, add the Sines of the aforefaid...
Side 231 - Perpendicular, let it fall from the End of a given Side, and oppofite to a given Angle...
Side 232 - The Tangents of the Bafes are proportional to the Tangents of the Angle at the Vertex.