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acute angle adjacent angle opposite angular Answers applications base called CHAPTER VII circle column common compasses complete Compute corresponding Cosec Cosine Cotang course cube decimal point diagram diameter difference direction distance Divide EDINBURGH employed equal EXAMPLE EXERCISES IN CHAPTER factors feet figures foot former FORMULÆ four fourth Geometry given given number given side greater half height horizontal hypothenuse inches integers interest latter length logarithms look marked measured miles minutes Multiply natural number nearly necessary NOTE number of degrees object observed operations Perp perpendicular plane position PRACTICAL preceding Price PROBLEM quadrant questions radius remove right-angled triangle RULE scale Secant seen sextant side Sine stands station Table taken Tang tangent telescope term third side tion Treatise Trigonometry turns unit vertex vertical angle yards
Página 284 - TO THEIR DIFFERENCE ; So IS THE TANGENT OF HALF THE SUM OF THE OPPOSITE ANGLES', To THE TANGENT OF HALF THEIR DIFFERENCE.
Página 247 - To find, then, by logarithms, the fourth term in a proportion, ADD THE LOGARITHMS OF THE SECOND AND THIRD TERMS, AND from the sum SUBTRACT THE LOGARITHM OF THE FIRST TERM.
Página 281 - From D as a center with a radius equal to a, draw an arc intersecting El in F and F'.
Página 294 - ... the angle of reflection is always equal to the angle of incidence, the image for any point can be seen only in the reflected ray prolonged.
Página 281 - Let abc (fig. 1 14) be a spherical triangle, whose sphere has its centre in o, and unity for radius. If now from c, on the plane aob, we let fall the perpendicular cd; from d on ae, bo, the perpendiculars de, df, and draw ce, cf; it would be easy to show that the triangles ceo, cfo are right angles...
Página 280 - To find a side, work the following proportion: — as the sine of the angle opposite the given side is to the sine of the angle opposite the required side, so is the given side to the required side.
Página 245 - BY LOGARITHMS. RULE. FROM the logarithm of the dividend subtract the logarithm of the divisor, and the number answering to the remainder will be the quotient required.