Chap. VIII.
GEOMETRY.
871
are charged with sights, and work on a circle described on the cross, which is divided into360 degrees. This alhidade serving to form angles has a magnetic needle at its centre,to mark the point of the compass, by placing one side of the square against the object to betaken. When the double cross has not an alhidade, four more sights are screwed in theintervals of the others, so that the cross has eight sights: it is to be remarked that largesquares are to be preferred to small, as they direct the visual rays better.
To ascertain whether the cross staff is accurate, mount it on its stand, and plant it on alevel piece of ground; then, boning by the sights AC, place a piquet with its card inthe line with the visual ray; and, without removing the cross staff, bone through thesights C, A, and plant a piquet H in the visual ray ; bone through all the other sights,and if they cut the same points, as well as the piquets placed at right angles, it may beconsidered correct.
To measure acute-angled Triangles. —Let fall from an angle of tbe triangle a perpendicularto the side opposite the angle, and multiply this perpendicularby tbe length of the side opposite the angle; half the productwill give the content or superficies of the triangle required.
To measure the acute-angled triangle ABC, plant piquets at.A B C, and place the cross staff at some point on theside B C, so that by looking along the sights of one diameter,the two piquets B and C may be seen, and without removingthe staff, look also at the piquet A, through the sights of theother diameter; if at D it cannot be seen, the staff is toofar to the right, and must be advanced to the left, as at K;then from the point E the two piquets B and C may be seenthrough the two sights of the diameter, which is in a line withB C, and through the other sights the piquet A ; measure theperpendicular E A, which is 24 feet, and BC 52: then mul-tiply the length EA, 80 feet, by BC, 52, and take half theproduct 4160, we have 2080 superficial feet for the content ofA 15 C, the acute-angled triangle.
To measure right-angled Triangles .—Multiply the two sideswhich form the right angle together, and half the product willbe the superficial content. Measure the side B C, which is 1*14feet 5 inches, and AB 125 feet; multiply 144 feet 5 inches by125 feet, and the product 18104 feet 6 inches divided gives9052 feet 3 inches for the superficial content. For if a parallel-ogram and triangle be upon the same base, and between thesame parallels, the parallelogram will be double the triangle,as shown by Euclid in his 41st prop. Book I. It is verymanifest that triangles are the halves of parallelograms upon thesame base, or upon equal bases and between the same parallels:and because these parallelograms are equal to one another, thetriangles which are their halves are also equal: on this accountit is said that the rectangle is equal to the product of its baseand altitude, and the triangle to half the product of its base andaltitude. When the two straight lines which form a rectangleare of such a length that one contains exactly the same number offeet or divisions that there are in the other, or side adjoining it;then the rectangle under these two straight lines is contained asmany times in the given rectangle as is expressed by the productof the two numbers, which denote how often the feet or divisionsare contained in the two sides.
To measure obtuse-angled Triangles. — After having plantedpiquets at the three angles A,B, C, place the cross staff at somepoint on the side BC, as at D, in such a manner that by look-ing along the sights of one diameter, the two piquets B andC are seen, and by those on the other diameter the piquet A ;if the piquet is not seen, the square is not exactly oppositethe angle A, and must be advanced to For E. Then measurethe perpendicular F A, which is 220 feet, and B C 663 feet,which multiplied produces 145860, the half of which, 72930,feet is the superficial content required.
To measure inaccessible Angles. — First find by the trigono-metric rules the length of the sides of the triangle, viz. A B68, BC 70, CA 28, and draw a similar triangle DHE;raise a perpendicular on the side D E, as H G, and proceed# as before described.
3 k 4
Fig. 1127.