zontal line I, between OO' and OB, is O'B, that is inch. Similarly the intercept on the horizontal line 2 is inch, on 3, inch, etc. Hence from p to q is one inch and seven-fortieths. In a similar manner diagonal scales can be made to divide any assumed unit-length into any required number of minute parts. The chief advantages of such scales are that the minute divisions are kept quite distinct and apparent, and that errors are consequently avoided. EXERCISES. 1. ABCD is a square and P is taken in BC so that PC is one-third of BC. AC cuts the diagonal BD in O, and Then OE is one-tenth of DB. AP cuts it in E. 2. If, in 1, OE is one-eighth of DB, how does P divide BC? 3. If BP is one nth of BC, what part of DB is OE? 4. Given three line-segments to find a fourth, so that the four may be in proportion. 5. The rectangle on the distances of a point and its chord of contact from the centre of a circle is equal to the square on the radius of the circle. 6. OD and DQ are fixed lines at right angles and O is a fixed point. A fixed circle with centre on OD and passing through O cuts OQ in P. Then OP. OQ is a constant however OQ be drawn. 7. To divide a given segment similarly to a given divided segment. 8. To divide a given segment into a given number of equal parts. 9. Two secants through A cut a circle in B, D, and C, E respectively. Then the triangles ABE and ACD are similar. So also are the triangles ABC and AED. 10. Two chords are drawn in a circle. To find a point on the circle from which perpendiculars to the chords are II. ABC is a triangle and DE is parallel to AC, D being on Then 12. If BO, in 11, cuts DE in P and AC in Q, BO is divided harmonically by P and Q. 13. A and B are centres of fixed circles and AX and BY are parallel radii. Show that XY intersects AB in a fixed point. 14. In the triangle ABC, BD bisects the B and cuts AC in D. Then BD2=AB. BC-AD.DC. (Employ the circumcircle.) 15. ABC is right-angled at B and BD is the altitude on AC. (1) The As ADB and BDC are each similar to ABC. (2) Show by proportion that AB2=AD. AC, 16. If R and r denote the radii of the circumcircle and incircle of a triangle, 2Rr(a+b+c)=abc. 17. In an equilateral triangle the square on the side is equal to six times the rectangle on the radii of the circumcircle and incircle. 18. OA, OB, OC are three lines. Draw a line cutting them so that the segment intercepted between OA and OC may be bisected by OB. 19. What is the measure of an angle in radians when its measure in degrees is 68° 17'? 20. How many radians are in the angle of an equilateral ▲? 21. The earth's diameter being 7,960 miles, what is the distance in miles between two places having the same longitude but differing 16° in latitude? 22. Construct a regular pentagon, a regular decagon, a regular polygon-of 15 sides, of 30 sides, of 60 sides. 23. ABCDE is a regular pentagon. (1) Every diagonal is divided into extreme and mean ratio by another diagonal. (2) The diagonals enclose a second regular pentagon. 24. Compare the side and the areas of the two pentagons of 23 (2). 25. If one side of a right-angled triangle is a mean proportional between the other side and the hypothenuse, the altitude from the right angle divides the hypothenuse into extreme and mean ratio. 26. A variable line from a fixed point A meets a fixed circle in P, and X is taken on AP so that AP.AX=a constant. The locus of X is a circle. 27. If two circles touch externally their common tangent is a mean proportional between their diameters. 28. Four points on a circle are connected by three pairs of lines. If a, a1 denote the perpendiculars from any fifth point on the circle to one pair of lines, ß, ß1 to another pair, and y, 1 to the third pair, then aa1=ßß1 =71(Employ 204°.) 29. A line is drawn parallel to the base of a trapezoid and bisecting the non-parallel sides. Compare the areas of the two trapezoids formed. 30. Draw two lines parallel to the base of a triangle so as to trisect the area. 31. ABC is right-angled at B, and AP is the perpendicular from A to the tangent to the circumcircle at B. Then AP.AC-AB2. SECTION II. FUNCTIONS OF ANGLES.-AREAL RELATIONS. 212°. Def.—When an element of a figure undergoes change the figure is said to vary that element. If a triangle changes into any similar triangle it varies its magnitude while its form remains constant; and if it changes into another form while retaining the same area, it varies its form while its area remains constant, etc. Similar statements apply to other figures as well as triangles. When a triangle varies its magnitude only, the tensors or ratios of the sides taken two and two remain constant. Hence the tensors or ratios of the sides of a triangle taken two and two determine the form of the triangle but not its magnitude; i.e., they determine the angles but not the sides. (77°, 3; 197°; 198°; 201°) A triangle, which, while varying its size, retains its form, is sometimes said to remain similar to itself, because the triangles due to any two stages in its variation are similar to one another. 213°. In the right-angled triangle the ratios or tensors of the sides taken in pairs are important functions of the angles and receive distinctive names. The AOPM is right-angled at M, and the POM is denoted by 0. P' Sup of M' P M Then, PM is the sine of e, and is contracted to sin 0 in writing. OM is the sine of the LOPM, but as LOPM is the comple ОР ment of 0, this tensor is called the cosine of 0, and is written cose. Cor. 2. or OP OM' ..tan 0 sin 0 cos Ꭿ 2 PM3 +OM2=OP2, .. (PM)2+(OM)2 = 1. sin20+cos20=1. 214°. Let OP=OP be drawn so that P'OM'=POM=0, and let P'M' be 1 on OM. Then P'OM is the supplement of 0, and the triangles P'OM' and POM are congruent. I. sin P'OM= P'M' PM i.e., an angle and its supplement have the same sine. 2. Cos P'OM= OM' But in changing from OM to OM', 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. an angle and its supplement have cosines which are equal in numerical value but opposite in sign. 215°. Theorem.-The area of a parallelogram is the product of two adjacent sides multiplied by the sine of their included angle. B (152°, 1) AC is a and BP is upon AD. Then BP is the altitude, and the area=AD.BP. α (153°, 1) =AB. AD sin BAP=ab sin 0. 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. 216°. Theorem.-The area of any quadrangle is one-half the product of the diagonals multiplied by the sine of the angle between them. ABCD is a quadrangle of which AC and BD are diagonals. Let LAOB=0=LCOD. Then LBOC=LAOD=supp. of 0. ▲AOB=OA. OB sin 0, B C D ABOC OB.CO sin 0, ACODOC. DO sin 0, |