But as CAD' is the supplement of the obtuse angle A, cos CAD' = cos A, and AD' = - b cos A. Either of these values, being substituted for AD, in (1), gives, a2 = b2 + c2 2bc cos A; whence, cos A = If we add 1 to both members, and recollect that 1+ cos A = 2 cos24 (Art. 66), Equation (4), we have, Substituting in (3), and extracting the square root, b+c a the plus sign, only, being used, since A < 90°; hence, The cosine of half of either angle of a plane triangle, is equal to the square root of half the sum of the three sides, into half that sum minus the side opposite the angle, divided by the rectangle of the adjacent sides. By applying logarithms, we have, log cos +4 = [logs + log (48 − a) + (a. c.) log b + (a. c.) log c]. If we subtract both members of Equation (2), from 1, Substituting in (5), and reducing, we have, The sine of half an angle of a plane triangle, is equal to the square root of half the sum of the three sides, minus one of the adjacent sides, into the half sum minus the other adjacent side, divided by the rectangle of the adjacent sides. Applying logarithms, we have, log sin A = 1 [log (18 b) + log (18 c) + (a. c.) log b + (a. c.) log c]. (B.) Third Case. To find the area of a triangle, when the But, from Formula (A'), Trig., Art. 66, we have, Find half the sum of the three sides, and from it subtract each side separately. Find the continued product of the half sum and the three remainders, and extract its square root; the result will be the area required. It is generally more convenient to employ logarithms; for this purpose, applying logarithms to the last equation, we have, [logs + log (48 − a) + log (†s — b) + log (ts—c)] hence, we have the following RULE. Find the half sum and the three remainders as before, then find the half sum of their logarithms; the number correspond ing to the resulting logarithm will be the area required. EXAMPLES. 1. Find the area of a triangle, whose sides are 20, 30, and 40. s = We have, 45, 48-a25, 8-b15, 8-c=5. By the first rule, = √45 × 25 × 15 × 5 290.4737 Ans. 2. How many square yards are there in a triangle, whose sides are 30, 40, and 50 feet? To find the area of a trapezoid. Ans. 661. 98. From the principle demonstrated in Book IV., Prop. VII., we may write the following RULE. Find half the sum of the parallel sides, and multiply it by the altitude; the product will be the area required. EXAMPLES. 1. In a trapezoid the parallel sides are 750 and 1225, and the perpendicular distance between them is 1540; what is the area? Ans. 1520750. 2. How many square feet are contained in a plank, whose length is 12 feet 6 inches, the breadth at the greater end 15 inches, and at the less end 11 inches ? Ans. 131. 3. How many square yards are there in a trapezoid, whose parallel sides are 240 feet, 320 feet, and altitude 66 feet? Ans. 2053 sq. yd To find the area of any quadrilateral. 99. From what precedes, we deduce the following RULE. Join the vertices of two opposite angles by a diagonal; from each of the other vertices let fall perpendiculars upon this diagonal; multiply the diagonal by half of the sum of the perpendiculars, and the product will be the area re quired. EXAMPLES. 1. What is the area of the quadrilateral ABCD, the diagonal ⚫AC being 42, and the perpendiculars Dg, Bb, equal to 18 and 16 feet? Ans. 714 sq. ft. 9 B 2. How many square yards of paving are there in the quadrilateral, whose diagonal is 65 feet, and the two perpendiculars let fall on it 28 and 33 feet? Ans. 222. To find the area of any polygon. 100. From what precedes, we have the following RULE. Draw diagonals dividing the proposed polygon into tra pezoids and triangles: then find the areas of these figures separately, and add them together for the area of the whole polygon. EXAMPLE. 1. Let it be required to determine the area of the polygon ABCDE, having five sides. Let us suppose that we have mea E sured the diagonals and perpendicu- Dd= |