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acute angles adjacent angles angle of elevation base line bearing calculation called centre chain circumference compass construction convenient cosecant cosine cotangent course and distance degrees determined diagonal difference of latitude difference of level difference of longitude direction Dist distance sailed divided draw equal to radius example extended feet find the angle find the remaining given side height horizontal plane hypothenuse instrument line of chords manner marked measure Merid meridian middle latitude miles minutes multiply observed parallel perpendicular plane sailing plane triangle plumb line principal meridian quadrant remaining sides represent right angled triangle river scale secant planes series of triangles ship in latitude side AC similar triangles sines and cosines square station subtracting survey tang tangent and secant theodolite townships triangle ABC fig vertical whence yards
Page 39 - ... the square of the hypothenuse is equal to the sum of the squares of the other two sides.
Page 59 - ... required the distance between the objects. Ans. 173.656 feet. 2. In order to find the distance between two trees, A and B, which could not be directly measured because of a pool which occupied the intermediate space, the distances of a third point C from each of them were measured, and also the included angle ACB: it was found that, CB = 672 yards, CA = 588 yards, required the distance AB.
Page 48 - The sum of any two sides of a triangle is to their difference, as the tangent of half the sum of the angles opposite to those sides, to the tangent of half their difference.
Page 59 - I measured out for a base 400 yards in a right line by the side of the river, and found that the two angles, one at each end of this line, subtended by the other end and the house, were 68° 2
Page 173 - Radius, is to the tangent of the course; as the meridional difference of latitude, is to the Difference of longitude.
Page 58 - Ex. 2. From the edge of a ditch 18 feet wide, surrounding a fort, the angle of elevation of the wall was found to be 62° 40'. Required the height of the wall, and the length of a ladder necessary to reach from my station to the top of it. Ans.
Page 46 - But a shorter way would be to take the arithmetical complement (Alg. 248) of the first logarithm and add it to the two others; thus, 0,17032 2,30103 9,86763 2,33898 the same as before. In the solution of this problem we have made use of the theorem, the sines of the angles are to each other as the sides opposite to these angles.
Page 59 - Wanting to know the breadth of a river, I measured a base of 500 yards in a straight line close by one side of it ; and at each end of this line I .found the angles subtended by the other end and a tree, close to the bank on the other side of the river, to be 53° and 79° 12'.
Page 17 - For this purpose it is divided into 360 equal parts called degrees, each degree into 60 equal parts called minutes, and each minute into 60 equal parts called seconds. The degrees, minutes, and seconds are marked thus ° ' " ; and 9° 18' 16", are read, 9 degrees 18 minutes and 16 seconds.