Using ruler and a pair of compasses only, (a) construct : (i) \(\Delta\)XYZ such that |XY| = 10cm, < XYZ = 30° and < YXZ = 45°. (ii) locus, \(l_{1}\), of po...

Construction using ruler and a pair of compasses only

(a)(i) Triangle XYZ. Draw the base \(|XY| = 10\text{ cm}\). At \(X\) construct \(\angle YXZ = 45^{\circ}\): first raise a perpendicular at \(X\) to obtain \(90^{\circ}\), then bisect that right angle to get \(45^{\circ}\). At \(Y\) construct \(\angle XYZ = 30^{\circ}\): first construct a \(60^{\circ}\) angle (an arc of radius equal to \(|XY_1|\) stepped off along the base), then bisect it to get \(30^{\circ}\). The ray from \(X\) and the ray from \(Y\) meet at \(Z\). This fixes \(\triangle XYZ\), and the third angle is \[\angle XZY = 180^{\circ} - 45^{\circ} - 30^{\circ} = 105^{\circ}.\]

(a)(ii) Locus \(l_1\). The set of points equidistant from \(Y\) and \(Z\) is the perpendicular bisector of \(YZ\). With centre \(Y\) and then centre \(Z\), using the same radius (a little more than half of \(|YZ|\)), draw arcs that cross above and below \(YZ\); the straight line through the two crossings is \(l_1\).

(a)(iii) Locus \(l_2\). The line through \(Z\) parallel to \(XY\). Since \(XY\) is the base, drop a perpendicular from \(Z\) to \(XY\) and at \(Z\) construct a right angle to that perpendicular; the resulting line through \(Z\) is parallel to \(XY\) and is \(l_2\).

(b) Mark \(M\), the point where \(l_1\) and \(l_2\) intersect.

The complete construction is shown below (measured values: \(|XZ| \approx 5.2\text{ cm}\), \(|YZ| \approx 7.3\text{ cm}\)).

(c) Measurement of \(\angle ZMY\). Because \(M\) lies on \(l_1\), it is equidistant from \(Y\) and \(Z\), so \(|MY| = |MZ|\) and \(\triangle MYZ\) is isosceles. On measuring the angle at \(M\) with a protractor: \[\boxed{\angle ZMY = 120^{\circ}.}\]