2. The side of a regular pentagon is 24 ft. Find quantities as in Ex. 1. 3. The side of a regular octagon is 24 ft. Find quantities as in Ex. 1. 4. The radius of a circle is 24 ft. Find the lengths of the sides and apothems of the inscribed regular triangle, quadrilateral, pentagon, hexagon, heptagon, and octagon. Compare the area of the circle and the areas of these regular polygons; also compare the perimeters of the polygons and the circumference of the circle. 5. For the same circle as in Ex. 4, find the lengths of the sides of the circumscribing regular figures named in Ex. 4. Compare their areas and perimeters with the area and circumference of the circle. 6. If a be the side of a regular polygon of n sides, show that R, the 180° radius of the circumscribing circle, is equal to a cosec -; and that r, the 180° radius of the circle inscribed, is equal to a cot n n 7. If r be the radius of a circle, show that the side of the regular inscribed polygon of n sides is 2r sin and that the side of the regular circum 180° n 180° -; scribing polygon is 2 r tan n 8. If a be the side of a regular polygon of n sides, R the radius of the circumscribing circle, and r the radius of the circle inscribed, show that area of polygonna2 cot· = nR2 sin 180° 360° = nr2 tan n 180° 34. Solution of oblique triangles. Since an oblique triangle can be divided into right-angled triangles by drawing a perpendicular from a vertex to the opposite side, it may be expected that knowledge concerning the solution of right-angled triangles will be of service in solving oblique triangles. This expectation will not be disappointed. An examination, which it is advisable for the student to make before proceeding farther, will show that all the sets of data from which a definite triangle can be drawn are those indicated in (1)–(4) below. The ability to make the following geometrical constructions is presupposed: (1) To draw a triangle on being given two of its angles and a side opposite to one of them; (2) To draw a triangle on being given two of its sides and an angle opposite to one of them; (3) To draw a triangle on being given two of its sides and their included angle; (4) To draw a triangle on being given its three sides. In what follows, only the steps in the solutions will be indicated. The examples that are worked may be saved, so that the amount of labor required by the method of solution shown here can be compared with the amount required by another method which will be described later. There are four cases in the solution of oblique triangles; these cases correspond to the four problems of construction stated above. CASE I. Given two angles and a side opposite to one of them. A A α с А B D FIG. 30. D B In ABC let A, B, a be known. Angle C and sides b, c are required. From C draw CD at right angles to AB or AB produced. In triangle CBD, angle CBD and side CB are known. .. BD and DC can be found. Then, in triangle ACD, side DC and angle A are known. .. AC and AD can be found. Side AB = AD – BD when B is obtuse, and AB = AD + DB when B is acute. Angle C 180° — (A + B). Another method of solution is given in Art. 55. EXAMPLES. 1. Ex. 1, Art. 55. 3. Ex. 2, Art. 60. 2. Ex. 2, Art. 55. 4. Other Exs. in Arts. 55, 60. CASE II. Given two sides and an angle opposite to one of them. N.B. The first part of the text in Art. 56 should be read at this time. Let (Fig. 30) AC, BC, angle A be known. [In a certain case, as shown in Art. 56, two triangles can be drawn which satisfy the given conditions.] From C draw CD at right angles to AB or AB produced. In ACD, AC and A are known. .. AD, DC, angle ACD, can be found. Then, in BCD, BC and CD are known. .. BD, angle DBC, can be found. In one figure, AB = AD – BD, angle ABC = 180° — CBD. angle ACB180° (CAB + ABC). Another method of solution is given in Art. 56. EXAMPLES. 1. Ex. 1, Art. 56. 3. Ex. 1, Art. 60. 2. Ex. 2, Art. 56. 4. Other Exs. in Arts. 56, 60. CASE III. Given two sides and their included angle. In ABC let b, c, A be known. Side a, B, C are required. From C draw CD at right angles to AB or AB produced. In ACD, AC and angle A are known. .. CD and AD can be found. Then, in triangle CDB, CD is now known, and BD=AD-AB or AB - AD. .. Angle CBD can be found. Angle ABC= 180° - CBD in figure on the left. Angle ACB · 180° − (A + B). Another method of solution is given in Art. 57. = In ABC let a, b, c be known. The angles A, B, C are required. From any vertex C draw CD at right angles to AB or AB pro Another method of solution is given in Art. 58. 1. Ex. 1, Art. 58. EXAMPLES. 2. Ex. 2, Art. 58. 3. Ex. 1, Art. 62. 4. Other Exs. in Arts. 58, 62. 5. Solve some of the problems in Art. 63 by means of right-angled triangles. 34 a. The area of a triangle in terms of the sides. (See Fig. 30.) From (1), (2), Case IV., Art. 34, Let then _ [2 cb + b2 + c2 − a2][2 cb — (b2 + c2 — a2)] = = (a + b + c)(− a +b+c) (a − b + c)(a + b −c). 4 c2 a+b+c=28; 2(sa) = a+b+c−2a=− a+b+c. Similarly, 2 (s – b) = a − b + c ; 2 (s − c) = a + b − c. CD = = √s(s − a) (s — b)(s — c). с .. Area ABC= } AB · CD = √s(s — a)(s — b)(s — c).* Ex. Find the areas of the triangles in Exs. Case IV., Art. 34. Check the results by finding the areas by the method of Art. 31. *This is sometimes known as Hero's Formula for the area of a triangle. It was discovered by Hero (or Heron) of Alexandria, who lived about 125 B.C., and placed engineering and land surveying on a scientific basis. 34 6. Distance and dip of the visible horizon. Let C be the centre of the earth, and let the radius be denoted by r. its height PL be denoted by h. Join P, C; draw PB from P to any point in the visible horizon; draw the horizontal line PH in the same plane with PC, PB. Then angle HPB is called the dip of the horizon. By geometry, angle PBC = 90°. PB' = PC2 — CB2 = (r + h)2 — r2 = 2 rh + h2. Since h2 is very small compared with 2 rh, Taker 3960 mi., and let h be measured in feet. Then of P in miles. P H Hence, the distance of the horizon in miles is approximately equal to the square root of one and one-half times the height in feet. EXAMPLES. 1. A man whose eye is 6 ft. from the ground is standing on the seashore. How far distant is his horizon? Distance V3 × 6 mi. = 3 mi. 2. Find the greatest distance at which the lamp of a lighthouse can be seen, the light being 80 ft. above the sea level. 3. Find the height of the lamp of a lighthouse above the sea level when it begins to be seen at a distance of 12 mi. 4. From the top of a cliff, 40 ft. above the sea level, the top of a steamer's funnel which is known to be 30 ft. above the water is just visible. is the distance of the steamer? What 5. Find the distance and dip of the horizon at the top of a mountain 3000 ft. high? 6. Find the distance and dip of the horizon at the top of a mountain 21 mi. high. 34 c. Examples in the measurement of land. In order to find the area of a piece of ground, a surveyor measures distances and angles sufficient to provide data for the computation. An account |