Given the three angles of a spherical triangle; required = log cosa [log cos (SB) + log cos (S—C) a. c. log sin B-a. c. log sin C]. MENSURATION. 156. Definition and Classification. Mensuration is the art of calculating the values of geometrical magnitudes. Mensuration is divided into two branches Mensuration of surfaces and Mensuration of volumes. MENSURATION OF SURFACES. 157. Unit of Superficial Measure. A unit of superficial measure is a square each side of which is a linear unit. Thus, according to the object to be accomplished, a square inch, a square foot, a square yard, an acre, etc., is the superficial unit taken. 158. Problem. To find the area of a rectangle. Let k denote the area, b the base, and a the altitude of a rectangle. There are a rows of b superficial units each. Ն Since there are b superficial units in one row, in a such rows there will be a times b or ab superficial units. The above demonstration applies only in case the base and altitude are commensurable, or have a common unit. If the base and altitude are incommensurable, denote the area by k', the base by b', and the altitude by a'. Then, since by Geometry any two rectangles are to each other as the products of their bases and altitudes, we have To find the area of a parallelogram. 1. When the base and altitude are given. Let k denote the area, b the base, and a the altitude of a parallelogram. Since a parallelogram is equal to a rectangle, having the same base and a altitude, and since the area of the rectangle is equal to the product of its base and altitude, the area of the parallelogram is equal to the product of its base and altitude. .'. (1) k ab. 2. When two sides and their included angle are given. A C To find the area of a triangle. 1. When the base and altitude are given. Since a triangle is one-half the parallelogram having the same base and altitude, we have for the triangle, (1) kab. b 2. When two sides and their included angle are given. Since a triangle is one-half the parallelogram, having an equal angle and equal adjacent sides, we have for the triangle, b (2) k 3. When two angles and a side are given. The third angle is equal to 180° minus the sum of the given angles. Let, then, the angles and the side. b be given. By the last case, we have B A kbe sin A. But sin B sin C: b : C, .. C= b sin C Substituting this value of c, we have (3) k b2 sin A sin C = 2 sin B 4. When two sides and an angle opposite one of In case of one or two solutions determined by article 72, find the value or values of C and B from the formulas, (5) k = √≥ p(pa) (p—b) (≥ p—c). 6. When the perimeter and angles are given. Let p be the perimeter, and A, B, and C the angles. B By article 98, (10), (11), (12), A C 1p2 tan 4 tan B tan CVp(pa) (pb) (pc). .. (6) k p2 tan tan B tan C. 7. When the perimeter and radius of the inscribed circle are given. B . ABC k, BOC=ar, AOC br, AOB = cr. = ... k = (a + b + c) r; but a + b + c = p. .. (7) k = 1 pr. 161. Examples. 1. Find the area of a triangle whose base is 75 ft., and altitude is 24 ft. Ans. 900 sq. ft. 2. Two sides of a triangle are 25 yds. and 30 yds., respectively, and their included angle is 50°; required the area. Ans. 287.2665 sq. yds. |