I. P'M' OP' sin P'OM= i.e., an angle and its supplement have the same sine. 2. Cos P'OM= But in changing from OM to OM', OM' OP' on the same line, OM vanishes and then reappears upon the opposite side of O. Therefore OM and OM' have opposite senses (156°), and if we consider OM positive, OM' is negative. OM' OM, and hence PM OP = sin 0. an angle and its supplement have cosines which are equal in numerical value but opposite in sign. B с 215°. Theorem.-The area of a parallelogram is the product of two adjacent sides multiplied by the sine of their included angle. (152°, 1) AC is a and BP is upon AD. Then BP is the altitude, and the area=AD. BP. (153°, 1) BP AB in ✩BAP. A But 7=AB. AD sin ✩BAP=ab sin 0. Let LAOB=0=LCOD. Then BOC=LAOD=supp. of 0. α ABCD is a quadrangle of which AC and BD are diagonals. P b Cor. 1. Since the area of a triangle is one-half that of the parallelogram on the same base and altitude, .. the area of a triangle is one-half the product of any two sides multiplied by the sine of the included angle. Or 2A=ab sin C = bc sin A= ca sin B. D 216°. Theorem.-The area of any quadrangle is one-half the product of the diagonals multiplied by the sine of the angle between them. B A E ▲AOB=OA. OB sin 0, ABOC=OB. CO sine, ACOD=OC. DO sin 0, ADOA A DO. OA sin 0, and adding, (compare 162°) 217°. BD being the altitude to AC in the ▲ABC, we have from 172°, 2, a2 = b2+c2 − 26. AD. AD AB cos A = c cos A, a2= b2+c2-2bc cos A. B B D A D C When B comes to B' the LA becomes obtuse, and cos A changes sign. (214°, 2) If we consider the cosine with respect to its magnitude only, we must write + before the term 2bc cos A, when A becomes obtuse. But, if we leave the sign of the function to be accounted for by the character of the angle, the form given is universal. Cor. 1. ABCD is a parallelogram. then Consider the ▲ABD, BD2=a2+b2 - 2ab cos 0. Next, consider the ▲ABC. Since LABC is the supplement of 0, and BC=AD=b, AC2=a2+b2 +2ab cos 0. and writing these as one expression, gives both the diagonals of any one of whose angles is ". 9 Cor. 2. But .. DE= =a cos 0 (CE being 1 to AD), CE=a sin 0. and a sin @ tan CAE b+a cos 0' which gives the direction of the diagonal. CE AE 218°. Def.-The ratio of any area X to another area Y is the measure of X when Y is taken as the unit-area, and is accordingly expressed as (Compare 188°.) X I. Let X and Y be two similar rectangles. Then X=ab and Y=a'b', where a and b are adjacent sides of the X and a' and ' those of the Y. But because the rectangles are similar, i.e., the areas of similar rectangles are proportional to the areas of the squares upon homologous sides. 2. Let X and Y be two similar triangles. Then X=ab sin C, Y=ba'b'sin C, because the triangles are similar, (197°) i.e., the areas of similar triangles are proportional to the areas of the squares upon homologous sides. P P' A'D'2' R' But X α b Χ ab a2 Y a'b' a'2' 3. Let X denote the area of the pentagon ABCDE, and Y that of the similar pentagon B A A'B'C'D'E'. Then = Q DC2 Q' D'C'2 and AD2 R AC2 = A'C'' PR b ď b E P (X) = R P+Q+R = AD AC DC D'C" X DC2 Y D'C'2 And the same relation may be proved for any two similar rectilinear figures whatever. .. the areas of any two similar rectilinear figures are proportional to the areas of squares upon any two homologous lines. (Y) (195°, 3) (207°) 4. Since two circles are always similar, and are the limits of two similar regular polygons, ... the areas of any two circles are proportional to the areas of squares on any homologous chords of the circles, or on line-segments equal to any two similar arcs. 5. When a figure varies its magnitude and retains its form, any similar figure may be considered as one stage in its variation. Hence the above relations, 1, 2, 3, 4, may be stated as follows: The area of any figure with constant form varies as the square upon any one of its line-segments. EXERCISES. 1. Two triangles having one angle in each equal have their areas proportional to the rectangles on the sides containing the equal angles. 2. Two equal triangles, which have an angle in each equal, have the sides about this angle reciprocally proportional, i.e., a: a' b': b. 3. The circle described on the hypothenuse of a right-angled triangle is equal to the sum of the circles described on the sides as diameters. 4. If semicircles be described outwards upon the sides of a right-angled triangle and a semicircle be described inwards on the hypothenuse, two crescents are formed whose sum is the area of the triangle. 5. AB is bisected in C, D is any point in AB, and the curves are semicircles. R S B 6. If a, b denote adjacent sides of a parallelogram and also of a rectangle, the ratio of the area of the parallelogram to that of the rectangle is the sine of the angle of the parallelogram. 7. The sides of a concyclic quadrangle are a, b, c, d. Then the cosine of the angle between a and b is (a2 + b2 − c2 − d2)/2(ab+cd). 8. In the quadrangle of 7, if s denotes one-half the perimeter, area={(sa)(s—b)(s— c)(s− d)}. 9. In any parallelogram the ratio of the rectangle on the sum and differences of adjacent sides to the rectangle on the diagonals is the cosine of the angle between the diagonals. Io. If a, b be the adjacent sides of a parallelogram and the angle between them, one diagonal is double the other -3 ( 2 + 2)2 II. If one diagonal of a parallelogram is expressed by the other diagonal is ʼn times as long. √(26072323 n2 + I when cos 0: = 12. Construct an isosceles triangle in which the altitude is a mean proportional between the side and the base. 13. Three circles touch two lines and the middle circle touches each of the others. Prove that the radius of the middle circle is a mean proportional between the radii of the others. 14. In an equilateral triangle describe three circles which shall touch one another and each of which shall touch a side of the triangle. 15. In an equilateral triangle a circle is described to touch the incircle and two sides of the triangle. its radius is one-third that of the incircle. Show that 4 M |