EXERCISES ON LINES AND ANGLES. (Graphical Exercises.) 1. Construct (with ruler and compasses only) an angle of 60°. By repeated bisection divide this angle into four equal parts. 2. By means of Exercise 1, trisect a right angle; that is, divide it into three equal parts. Bisect each part, and hence shew how to trisect an angle of 45°. [No construction is known for exactly trisecting any angle.] 3. Draw a line 6.7 cm. long, and divide it into five equal parts. Measure one of the parts in inches (to the nearest hundredth), and verify your work by calculation. [1 cm. =0.3937 inch.] 4. From a straight line 3.72" long, cut off one seventh. Measure the part in centimetres and the nearest millimetre, and verify your work by calculation. 5. At a point X in a straight line AB draw XP perpendicular to AB, making XP 18" in length. From P draw an oblique PQ, 30' long, to meet AB in Q. Measure XQ. (Problems. State your construction, and give a theoretical proof.) 6. In a straight line XY find a point which is equidistant from two given points A and B. When is this impossible? 7. In a straight line XY find a point which is equidistant from two intersecting lines AB, AC. When is this impossible? 8. From a given point P draw a straight line PQ, making with a given straight line AB an angle of given magnitude. 9. From two given points P and Q on the same side of a straight line AB, draw two lines which meet in AB and make equal angles with it. [Construction. From P draw PH perp. to AB, and produce PH to P', making HP' equal to PH. Join P'Q cutting AB at K. Join PK. Prove that PK, QK are the required lines.] 10. Through a given point P draw a straight line such that the perpendiculars drawn to it from two points A and B may be equal. Is this always possible? THE CONSTRUCTION OF TRIANGLES. PROBLEM 8. To draw a triangle having given the lengths of the three sides. Let a, b, c be the lengths to which the sides of the required triangle are to be equal. Construction. Draw any straight line BX, and cut off from it a part BC equal to a. With centre B, and radius c, draw an arc of a circle. With centre C, and radius b, draw a second arc cutting the first at A. Join AB, AC. Then ABC is the required triangle, for by construction the sides BC, CA, AB are equal to a, b, c respectively. Obs. The three data, a, b, c may be understood in two ways either as three actual lines to which the sides of the triangle are to be equal, or as three numbers expressing the lengths of those lines in terms of inches, centimetres, or some other linear unit. NOTES. (i) In order that the construction may be possible it is necessary that any two of the given sides should be together greater than the third side (Theorem 11); for otherwise the arcs drawn from the centres B and C would not cut. (ii) The arcs which cut at A would, if continued, cut again on the other side of BC. Thus the construction gives two triangles on opposite sides of a common base. ON THE CONSTRUCTION OF TRIANGLES. It has been seen (Page 50) that to prove two triangles identically equal, three parts of one must be given equal to the corresponding parts of the other (though any three parts do not necessarily serve the purpose). This amounts to saying that to determine the shape and size of a triangle we must know three of its parts: or, in other words, To construct a triangle three independent data are required. For example, we may construct a triangle (i) When two sides (b, c) and the included angle (A) are given. The method of construction in this case is obvious. (ii) When two angles (A, B) and one side (a) are given. Here, since A and B are given, we at once know C; (iii) If the three angles A, B, C are given (and no side), the problem is indeterminate, that is, the number of solutions is unlimited. For if at the ends of any base we make angles equal to B and C, the third angle is equal to A. This construction is indeterminate, because the three data are not independent, the third following necessarily from the other two PROBLEM 9. To construct a triangle having given two sides and an angle opposite to one of them. Let b, c be the given sides and B the given angle. Construction. Take any straight line BX, and at B make the XBY equal to the given B. From BY cut off BA equal to c. With centre A, and radius b, draw an arc of a circle. If this arc cuts BX in two points C, and C2, both on the same side of B, both the ABC1, ABC, satisfy the given conditions. This double solution is known as the Ambiguous Case, and will occur when b is less than c but greater than the perp. from A on BX, EXERCISE. Draw figures to illustrate the nature and number of solutions in the following cases: (i) When b is greater than c. (ii) When b is equal to c. (iii) When b is equal to the perpendicular from A on BX. (iv) When b is less than this perpendicular. |