(a) Using a ruler and a pair of compasses only, construct: (i) a triangle ABC such that |AB| = 5cm, |AC| = 7.5cm and < CAB = 120°; (ii) the locus \(l_{1}\) ...

Answer Details

(a) Construction (ruler and a pair of compasses only, drawn to scale 1 cm to 1 cm):

Steps.

1. Draw a base line and mark \(|AB| = 5\text{ cm}\).
2. Construct \(\angle CAB = 120^{\circ}\) at \(A\). With centre \(A\) draw an arc cutting \(AB\); step the radius round the arc twice to obtain \(60^{\circ}\) and then \(120^{\circ}\), and rule the arm \(AX\) through the \(120^{\circ}\) mark.
3. With centre \(A\) and radius \(7.5\text{ cm}\) cut arm \(AX\) at \(C\), so \(|AC| = 7.5\text{ cm}\). Join \(BC\); triangle \(ABC\) is complete.
4. Locus \(l_{1}\): the perpendicular bisector of \(AB\). With centre \(A\), then centre \(B\), and the same radius (a little more than half of \(|AB|\)), draw arcs that meet above and below \(AB\); the line through the two meeting points is \(l_{1}\), the locus of points equidistant from \(A\) and \(B\).
5. Locus \(l_{2}\): the bisector of \(\angle CAB\). With centre \(A\) draw an arc cutting both \(AB\) and \(AC\); from those two cuts draw equal arcs meeting inside the angle, and rule \(l_{2}\) from \(A\) through that meeting point. \(l_{2}\) is the locus of points equidistant from lines \(AB\) and \(AC\).

(b) \(l_{1}\) and \(l_{2}\) intersect at the point marked \(P\) inside the triangle.

(c) Measurement. Measuring directly from the construction, \(|CP| = 6.7\text{ cm}\) (accept \(6.5\text{ cm}\) to \(6.7\text{ cm}\)).

This agrees with a coordinate check. Taking \(A(0,0)\) and \(B(5,0)\), then \(C = 7.5(\cos120^{\circ},\ \sin120^{\circ}) = (-3.75,\ 6.50)\). The perpendicular bisector of \(AB\) is \(x = 2.5\), and the bisector of the \(120^{\circ}\) angle rises at \(60^{\circ}\), so \(P = (2.5,\ 2.5\tan60^{\circ}) = (2.5,\ 4.33)\). Hence