Ans. A = 48° 31′ 18", B = 62° 55′ 44′′, C = 125° 18′ 56′′. CASE VI. The three angles being given, to find the sides. 90. The solution in this case is entirely analogous to the preceding one. Applying logarithms to Formula (2), Art. 82, we have, [log cos (SB) + log cos (S — C) In the same manner as before, we change the letters, to suit each case. EXAMPLES. 1. Given A 48° 30′, B = 125° 20′, and C = 62° 54', Ans. a = 56° 39′ 30′′, b = 114° 29′ 58′′, C = 83° 12' 06" 2. Given A = 109° 55′ 42′′, B = 116° 38′ 33′′, and C = 120° 43′ 37′′, to find a, b, and C. Ans. a = 98° 21′ 40′′, b = 109° 50′ 22′′, c = 115° 13′ 28′′. MENSURAΤΙΟΝ. 91. MENSURATION is that branch of Mathematics which treats of the measurement of Geometrical Magnitudes. 92. The measurement of a quantity is the operation of finding how many times it contains another quantity of the same kind, taken as a standard. This standard is called the unit of measure. 93. The unit of measure for surfaces is a square, one of whose sides is the linear unit. The unit of measure for volumes is a cube, one of whose edges is the linear unit. If the linear unit is one foot, the superficial unit is one square foot, and the unit of volume is one cubic foot. If the linear unit is one yard, the superficial unit is one square yard, and the unit of volume is one cubic yard. 94. In Mensuration, the term product of two lines, is used to denote the product obtained by multiplying the number of linear units in one line by the number of linear units in the other. The term product of three lines, is used to denote the continued product of the number of linear units in each of the three lines. Thus, when we say that the area of a parallelogram is equal to the product of its base and altitude, we mean that the number of superficial units in the parallelogram is equal to the number of linear units in the base, multiplied by the number of linear units in the altitude. In like manner, the number of units of volume, in a rectangular parallelopipedon, is equal to the number of superficial units in its base oultiplied by the number of linear units in its altitude. azd SO on. 95. MENSURATION OF PLANE FIGURES. To find the area of a parallelogram. From the principle demonstrated in Book IV., Prop. V., we have the following RULE. Multiply the base by the altitude; the product will be the area required. EXAMPLES. 1. Find the area of a parallelogram, whose base is 12.25, and whose altitude is 8.5. Ans. 104.125. 2. What is the area of a square, whose side is 204.3 feet? 3. How many square yards are whose base is 66.3 feet, and altitude Ans. 41738.49 sq. ft. there in a rectangle 33.3 feet? Ans. 245.31 sq. yd. 4. What is the area of a rectangular board, whose length is 12 feet, and breadth 9 inches ? 9 sq. ft 5. What is the number of square yards in a parallelo. gram, whose base is 37 feet, and altitude 5 feet 3 inches? Ans. 21. To find the area of a plane triangle. 96. First Case. When the base and altitude are given. From the principle demonstrated in Book IV., Prop. VI., we may write the following RULE. Multiply the base by half the altitude; the product will be the area required. EXAMPLES. 1. Find the area of a triangle, whose base is 625, and altitude 520 feet. Ans. 162500 sq. ft. 2. Find the area of a triangle, in square yards, whose base is 40, and altitude 30 feet. Ans. 663. 3. Find the area of a triangle, in square yards, whose base is 49, and altitude 25 feet. Ans. 68.7361. Second Case. When two sides and their included angle are given. Let ABC represent a plane tri angle, in which the side. BC= a, and the angle given. From A draw AD АВ = c, B, are perpen dicular to BC; this will be the altitude of the triangle. From For B D mula (1), Art. 37, Plane Trigonometry, we have, AD c sin B. Denoting the area of the triangle by Q, and applying the rule last given, we have, log (20) = log a + log c + log sin B-10; hence, we may write the following RULE. Add together the logarithms of the two sides and the Logarithmic sine of their included angle; from this sum subtract 10; the remainder will be the logarithm of double the area of the triangle. Find, from the table, the number answering to this logarithm, and divide it by 2; the quotient will be the required area. EXAMPLES. 1. What is the area of a triangle, in which two sides a and b, are respectively equal to 125.81, and 57.65, and whose included angle C, is 57° 25'? Ans. 2Q 6111.4, and = Q = 3055.7 Ans 2. What is the area of a triangle, whose sides are 30 and 40, and their included angle 28° 57'? Ans. 290.427. 3. What is the number of square yards in a triangle, of which the sides are 25 feet and 21.25 feet, and their included angle 45° ? Ans. 20.8694. LEMMA. To find half an angle, when the three sides of a plane triangle are given. 97. Let ABC be a plane tri- C angle, the angles and sides being de noted as in the figure. We have (B. IV., P. XII., XIII.), a2 = b2 + c2 2c. AD. a D' A De (1.) When the angle A is acute, we have (Art. 37), AD = cos A; when obtuse, AD' b cos CAD'. |