Given, the hypothenuse, H, and the sum of the legs, B, P (which call s) thence to find the legs respectively. Given, the hypothenuse, H, and the difference of the other sides, d, to find the sides B and P. = Given, the difference d, of the legs B, P, and the perpendicular p, demitted from the right angle to the hypothenuse; to find the hypothenuse. H B p+ √ d2 + p2=H. If, in a right-angled triangle, one of the acute angles be 30°, the side opposite that angle will be equal to half the hypothenuse: for, Let ABC be the triangle given, right angled at B, and the CAB=30°, then BC= AC 2 B K D DEMONSTRATION. Produce CB to D, so that BD may be equal to CB, and drawing AD, the two triangles I and K will be similar and equal; therefore, the CAD will be = 60°, and the triangle CAD will be equilateral; therefore the line CB, which is half of CD, will also be half of AC. PART IV. THE APPLICATION OF TRIGONOMETRY TO DISTANCES, &c. By the mensuration of one or more lines, and by taking the angle which one object bears towards another, we find out, trigonometrically, the relative distances between such objects as may be inaccessible by impediments of water, wood, or other obstruction. The instrument which a land measurer ought to choose, for taking the angles of bearing of one object to another, is the theodolite.* And for such lengths as he may have occasion to measure, he will, doubtlessly, employ a Gunter's chain†; as being best calculated for accuracy and dispatch. *For a description of this instrument see page 191. The measuring chain is composed of 100 links of strong iron wire, each link 7.92 inches; consequently, the whole chain is 22 yards or 4 poles, in length. At every 10 links is fastened a piece of brass, so cut into points as to denote the number of tens. The better to carry on the account of the number of chains, there are ten (or rather eleven) small iron arrows, 16 inches in length, used by the assistant who has the foremost end of the chain, one of which he sticks into the ground at the end of each chain's length; which are collected by the hindermost chainman, or master-measurer. EXAMPLE I. In the triangle ABC, given the A= 60°, the B=80°, and the base AB=500 links. The distance from A to C (inaccessible on account of wood) is required. By subtracting the sum of the angles A and B from 180°, the C is found 40°. = 1. In the foregoing triangle, given as before, required, by a scale of 6 chains to an inch, and also by logarithms, the side BC, inaccessible on account of water. Ansr. 673-7 links. 2. Supposing, also, / B was 85° 30' required the distance BC. Ansr. 764.5 links. EXAMPLE II. On the side of a river, whose breadth is required, suppose two stations taken at A and B, the distance from each other being 450 links; and the BAC 53°, and the CBA 60°, the perpendicular CP is required. = = By subtraction of A+B from 180, the ACB is found 67°. Then, by Case I. in Trigonometry, First, As s. ACB = 67° 9.964026 2-653212 9.937531 12.590743 2-626717 |