PROBLEM. Given the anglas and a leg of a right-angled triangle ; to find the other leg and the hypotentise. In the plane triangle ABC, right-angled at B, { '} } . Required AC and BC. Geometrically. Make AB=162, and the angle A=53° 07' 48"; then raise the perpendicular BC meeting AC in C. So shak AC measure 270, and BC 216.' Arithmetically. th to * DEMONSTRATION. With the centre A and any radius AD, describe an arc DE, and erect the perpendicular DF; which, it is evident, will be the Langent, and AF the secant of the arc DE, or angle A, to the radius AD. And in the similar triangles ADF, ABC, it will be AD : AB :: DF : BC :: AF : AC. Q. E. D. Arithmetically. 10'0000000 22095150 :: tang. of 53° 7' 48" 10'1249371 Instrumentally The extent from 45° to 53° 08', upon the tangents; will reach from 162 to 216 upon the numbers, NOTE 2. It is common to add another method for right-angled F triangles, which is this. ABC being the triangle, make a leg AB radius, that is, with centre A and radius AB, describe A an arc BF: then it is evident, that the other leg BC represents the tangent, and the hypotenuse AC the secant of the angle A or arc BF. In like manner, if the leg BC be made radius, then the leg AB will represent the tangent, and AC the secant of the arc BG, or the angle C. But But if the hypotenuse be made radius, then each leg will represent the sine of its opposite angle ; namely, the leg AB the sine of the arc AE or angle C, and the leg BC the sine of the arc CD or angle A. And then the general rule for all these cases is this ; the sides are to each other, as the parts, which they represent, END OF PLANE TRIGONOMETRI. MENSURATION OF SUPERFICIES. HE AREA of any figure is the measure of its surface, ; or the space contained within the bounds of the sure face, without any regard to thickness. The area is estimated by the number of squares contained in the surface, the side of those squares being either an inch, a foot, a yard, &c. And hence the area is said to be so many square inches, or square feet, or square yards, &c. Our ordinary lineal measures, or measures of length, are as in the first of the following tables ; and the annexed table of square measures is taken from it by squaring the several numbers. . SQUARE MEASURES. I foot i fathom I pole or rod I furlong 64 furlongs I mile. 9 feet 36 feet . PROBLEM I. a To find the area of a parallelogrom ; whether it be a squareg a rectangle, a rhombus, or a rhomboid. RULE.* Multiply the length by the breadth, or perpendicular height, and the product will be the area. EXAMPLES * Take any rectangle ABCD, and divide each of its sides into as many equal parts as is expressed by the number of times they contain the linear measuring unit, and let all the opposite poiots of division be connected by right lines. Then, it is evident, that these lines divide the rectangle into a number of squares, each equal to the superficial measuring unit, and that the number of these squares, or the area of the figure, is equal to the number of linear measuring units in the length as often repeated, as there are linear measuring units in the breadth or height, that is, equal to the length multiplied by the height, which is the rule. And since a rectangle is equal to an oblique parallelogram standing upon the same base, and between the same parallels ; (Erc., I. 35) therefore the rule is true for any parallelogram in general . Q. E. D. RULE |