Construct a triangle ABC with AB = 5.5 cm, AC = 6 cm and ∠BAC=105∘. Mark the point of intersection of the locus of points equidistant from BA and BC and the locus of points equidistant from B and C, as P. Then the length of PC is
4.8 cm
Steps of construction:
Δ ABC is the required triangle.
(We know that the locus of a point which is equidistant from two intersecting straight lines is a pair of straight lines which bisect the angles between the given lines. We draw the bisector of the angle B so as to get the locus of the point which is equidistant from BA and BC).
(We know that the locus of a point which is equidistant from two fixed points is the perpendicular bisector of the line segment joining the two fixed points. We draw the perpendicular bisector of BC so as to get the locus of the point which is equidistant from B and C).
From the construction, we see that locus of points equidistant from BA and BC and the locus of points equidistant from B and C intersect at P.
On measuring, we get, PC = 4.8 cm