New American Practical NavigatorU.S. Government Printing Office, 1821 |
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Strona 2
... hence there is one decimal in the quotient . EXAMPLE II . Divide 0.35 by 0.7 .7 ) .35 ( .5 .35 EXAMPLE III . Divide 3.1 by .0062 Previous to the division I affix a number of ciphers to the right hand of 3.1 , which does not alter its ...
... hence there is one decimal in the quotient . EXAMPLE II . Divide 0.35 by 0.7 .7 ) .35 ( .5 .35 EXAMPLE III . Divide 3.1 by .0062 Previous to the division I affix a number of ciphers to the right hand of 3.1 , which does not alter its ...
Strona 7
... Hence it is evident that all the angles which can be made from a point in any line , towards one side of the line , are equal to two right angles , and that all the angles which can be made about a point , are equal to four right angles ...
... Hence it is evident that all the angles which can be made from a point in any line , towards one side of the line , are equal to two right angles , and that all the angles which can be made about a point , are equal to four right angles ...
Strona 8
... Hence BEF + EFD = two_right angles ; and in the same manner we may prove AEF + CFE = two right angles . XXXIII . In any triangle ABC , one of its legs , as BC being produced towards D , the external angle ACD is equal to the sum of the ...
... Hence BEF + EFD = two_right angles ; and in the same manner we may prove AEF + CFE = two right angles . XXXIII . In any triangle ABC , one of its legs , as BC being produced towards D , the external angle ACD is equal to the sum of the ...
Strona 10
... Hence all angles ACB , ADB , AEB , & c . at the circum- ference of a circle standing on the same chord AB are equal to each other ; for they are all measured by the same arch , viz . half the arch AB . XLII . An angle in a segment ...
... Hence all angles ACB , ADB , AEB , & c . at the circum- ference of a circle standing on the same chord AB are equal to each other ; for they are all measured by the same arch , viz . half the arch AB . XLII . An angle in a segment ...
Strona 11
... hence ( by art . 37 ) BD is equal to AC , and the angle DAC equal to the angle ADB ; therefore ( by art . 34 ) the lines BD , AC , must be parallel . Cor . Hence it follows that the quadrilateral ABDC is a parallelogram , since the ...
... hence ( by art . 37 ) BD is equal to AC , and the angle DAC equal to the angle ADB ; therefore ( by art . 34 ) the lines BD , AC , must be parallel . Cor . Hence it follows that the quadrilateral ABDC is a parallelogram , since the ...
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2Cor a-back a-head a-lee Aldebaran anchor angle apparent altitude arch azimuth bearing and distance brace cable calculated Cape centre Co-secant Co-sine Co-tang column compass Corr correction corresponding course and distance degrees Degs Diff difference of latitude difference of longitude Dist equal error EXAMPLE feet Funchal given Greenwich haul head sails heave horizon glass HourA.M larboard Latitude and Departure lee-way line of numbers logarithm mast mean meridian meridian altitude method middle latitude miles mizen moon moon's multiplied N.sine Nautical Almanac nearly noon observed altitude parallax parallel perpendicular plane Plane Sailing radius refraction rope rule sails sea account Secant semi-diameter sextant ship ship's side sine square star star's staysail subtracted sun's declination sun's right ascension tack taken Tangent tide topsails TRAVERSE TABLE triangle true distance tude variation veer wind windward zenith distance
Popularne fragmenty
Strona 2 - In any triangle, the sum of the three angles is equal to two right angles, or 180°.
Strona 104 - ... or taking their difference when of contrary names ; the altitude to be reckoned from the south point of the horizon, when the latitude is north, and the contrary when south ; but when the sum exceeds 90°, it is to be taken from 180°...
Strona 166 - To find the solidity of a pyramid and of a cone. RULE. — Multiply the area of the base by one third of its altitude, and the product will be its solidity.
Strona 185 - The cause of the. tides is the unequal attraction of the sun and moon upon different parts of the earth. . For they attract the parts of the earth's surface nearest to them, with a greater force than they do its centre : and attract the centre more than they do the opposite surface. To restore this equilibrium the waters take a spheroidal figure, whose longer axis is directed towards the attracting luminary.
Strona 24 - To find the logarithm of a vulgar fraction. RULE. Subtract the logarithm of the denominator from the logarithm of the numerator...
Strona 186 - ... the miles the current sets per hour, and the bearing of the log will show the set of it. There is a very remarkable current, called the GULF STREAM, which sets in a north-east direction along the coast of America, * From Nathaniel Bowditch, THE NEW AMERICAN PRACTICAL NAVIGATOR, E.
Strona 9 - TO THEIR DIFFERENCE ; So IS THE TANGENT OF HALF THE SUM OF THE OPPOSITE ANGLES', To THE TANGENT OF HALF THEIR DIFFERENCE.
Strona 292 - In spherical triangles, whether right angled or oblique angled, the sines of the sides are proportional to the sines of the angles opposite to them.
Strona 167 - If the vessel be double-decked, take the length thereof from the fore part of the main stem, to the after part of the stern post, above the upper deck ; the breadth thereof at the broadest part above the main wales...