17. Wishing to find the height of the tower SD, I took a base line PB in the same hori zontal plane with the bottom m D of the tower, and placed stations at P, G, and B, making PG GBα. At P I found the angle of elevation = P; at G the angle of elevation = /G; and at в the angle of elevation B: required an expression for the height of the tower. = Let x SD; then sp=x cot P, SG=x cot G, and SB=x cot B ; therefore by Art. 51. Geo., x2 cot2p+ x2 cot2 B=2 a2+ 2 x2 cot2G; 18. At the top of a tower a feet high, the angles of depression of the top and bottom of an upright column, standing on the same level plane as the tower, were found to be 30° and 60°; required the height of the column. Ans. 2 a 3' 19. From the top of the Peak of Teneriffe, h miles high above the level of the sea, the visible horizon appeared depressed A°; required the radius of the earth. Ans. h. cot. cot A. 140 MENSURATION OF SURFACES. 1. PROBLEM. To find the area of a square, rectangle, or any parallelogram. RULE. Multiply the base by the perpendicular height or breadth, and the product will be the area. (See Geo. pages 31 and 34.) Note. The dimensions of artificers' work are usually taken in feet and inches. Land is measured by the chain, which is 66 feet long, and contains 100 links. An acre contains 10 sq. chains, or 100,000 sq. links. The content of land is also expressed in roods and perches: 40 perches make 1 rood, and 4 roods make 1 acre. 2. Required the area of a parallelogram ABCD, whose base AB is 3 feet 4 inches, and perpendicular height DE is 2 feet 8 inches. Here, as an inch is the 12th of a foot, we calculate by the scale of twelfths; observing that units multiplied by units give units, units multiplied by twelfths give twelfths, and twelfths multiplied by twelfths give 144ths, or 1 sq. inch. The 12ths are written with an accent over them, and the 144ths with two accents. 3. A door is 7 ft. 3 in. long, and 3 ft. 6 in. broad; required its area, and cost at 2s. 3d. per sq. foot. = By fractions. 7 ft. 3 in.7 ft., and 3 ft. 6 in. 31 ft. .. Area 71 x 3203-25 sq. ft. = .. Cost=2s. 3d. × 253 = £2 17s. 1d. 4. Required the area of a rectangular table 9 ft. 6 in. long, and 4 ft. 3 in. broad. Ans. 40 ft. 4′ 6′′. 5. Required the cost, in the last example, at 1s. 6d. per sq. foot. Ans. £3 Os. 6d. 6. What is the area of a square table, the side of which measures 3 ft. 8 in.? Ans. 13 ft. 5′ 4′′. 7. A rectangular floor is 29 ft. 2 in. long, and 18 ft. 4 in. broad; what did it cost in paving at 6s. 9d. per sq. yd.? Ans. £20 1s. 01⁄2d. 8. What is the area of a parallelogram, whose base is 6 ft. 4 in. and perpendicular height 3 ft. 5 in.? Ans. 21 ft. 7' 8". 9. Required the acreage of a field, in the form of a parallelogram, whose length or base is 930 links, and perpendicular breadth 480 links. Ans. 4 ac. 1 r. 34 p. 10. How many yards of carpet, 3 qrs. wide, will cover a room that measures 14 ft. 3 in. by 9 ft. 4 in.? Ans. 19 yds. 2 ft. 41⁄2 in. 11. How many square yards are there in a parallelogram whose length is 37 ft., and perpendicular breadth 2·625 ft.? Ans. 10.79. 12. How many acres are in a rectangular field, the length of which is 1375 links, and breadth 95 links? 13. Required the cost of a piece of ground 24 ft. 3 in. by 7 ft. 6 in., at 3s. 6d. per sq. foot. Here to find the surface in the frame, we shall subtract the area of the rectangle KLMN from ABCD. D Ans. £31 16s. 6d. N M K L Area rectangle ABCD=20 × 16=320 sq. in. Now KL 20–614, and KN=16-6=10; = .. Area rectangle KLMN= 14 × 10 140 sq. in. 15. A coach road goes round the rectangular field ABCD; it is required to find the area of the ground taken up by the coach road, when its breadth is 4 yds.; the length AB of the field being 88 yds., and the breadth 40 yds. Ans. 960 sq. yds. 2. PROBLEM. To find the area of a triangle when its base and perpendicular height are given. RULE. Multiply the base by the and half the product will be the area. EXAMPLES. perpendicular height, See Geo. p. 35. 1. Required the number of acres in the triangular field ABC, whose base AB is 945 links, and perpendicular CD 480 links. C 2. Required the area of a triangular field, one of whose sides measures 618 links, and the perpendicular on it, from the opposite angle, 730 links. Ans. 2 ac. 1 r. ·912 p. 3. How many square yards of brick-work are in the gable top of a house, whose breadth is 20 ft. 6 in., and perpendicular height 10 ft. 4 in.? Ans. 11 yds. 6 ft. 11'. 4. How many square yards are there in a right-angled triangle, whose base is 49 ft., and perpendicular 25-25 ft.? Ans. 68.736. 3. PROBLEM. To find the area of a triangle, given the three sides. RULE. From half the sum of the three sides subtract each side severally; multiply the half sum and the three re |