Can somebody plz give me the steps of construction for the following sum:- ? Construct ?ABC in which BC = 4.2 cm, ?ABC = 60 Solution Draw AX of sufficient length. Taking 5cm radius in compass, cut off AX at B . Now AB =5cm. Take the compass with 5cm and at A as centre, make an arc and similarly with the same radius in compass and B as centre and make arc. The two arc cut at a point P. Draw BP and mark off C on BP such that BC = 4.2 cm. Now Angle ABC = 60 degree. Bisect the angle ABC : With B as centre and with suitable radius,draw an arc to cut BA and BC at X and Y respectively. With X as centre and then Y as centre , draw arcs of equal radius to cut at Z. Draw BZ , the angle bisector of angle ABC. Extend BZ to a sufficient length. Mark off BO along BZ, such that BO = 4cm. Draw a circle with on BO as diameter with the centre L , the mid point of BO with radius LO= LB =2cm. The circle cuts the BA and BC at Points M and N. Then OM = ON should be equal to 2cm. And with O as centre you can draw a circle of radius OM =ON = 2cm , which touch both AB and BC at M and N. Proof: Since BZ is bisector of angle ABC, Angle ABZ = Angle ZBC = 30 degree. Triangle OMB = is a right angled triangle as it is in a semimicirlce. Therefore, angle OMB = 90 degree. Angle OBM= angle OBA = 30 degree. Therefore, in a right angled triangle , where one angle is 30 degree, the opposite side OM must be half of hypotenuse BO. So OM =2cm. Since angle BMO =90 degree, BA is the tangent to the circle of radius 2cm with centre O and BA touches the circle at M. Similar explanation could be given to say that BC is the tangent to the circle touching at N..

1. Can somebody plz give me the steps of construction for the following sum:- ? Construct ?ABC
in which BC = 4.2 cm, ?ABC = 60
Solution
Draw AX of sufficient length. Taking 5cm radius in compass, cut off AX at B . Now AB =5cm.
Take the compass with 5cm and at A as centre, make an arc and similarly with the same radius in
compass and B as centre and make arc. The two arc cut at a point P. Draw BP and mark off C on
BP such that BC = 4.2 cm. Now Angle ABC = 60 degree.
Bisect the angle ABC : With B as centre and with suitable radius,draw an arc to cut BA and BC
at X and Y respectively.
With X as centre and then Y as centre , draw arcs of equal radius to cut at Z. Draw BZ , the angle
bisector of angle ABC. Extend BZ to a sufficient length.
Mark off BO along BZ, such that BO = 4cm.
Draw a circle with on BO as diameter with the centre L , the mid point of BO with radius LO=
LB =2cm.
The circle cuts the BA and BC at Points M and N.
Then OM = ON should be equal to 2cm. And with O as centre you can draw a circle of radius
OM =ON = 2cm , which touch both AB and BC at M and N.
Proof: Since BZ is bisector of angle ABC, Angle ABZ = Angle ZBC = 30 degree. Triangle OMB
= is a right angled triangle as it is in a semimicirlce. Therefore, angle OMB = 90 degree. Angle
OBM= angle OBA = 30 degree. Therefore, in a right angled triangle , where one angle is 30
degree, the opposite side OM must be half of hypotenuse BO. So OM =2cm. Since angle BMO
=90 degree, BA is the tangent to the circle of radius 2cm with centre O and BA touches the circle
at M. Similar explanation could be given to say that BC is the tangent to the circle touching at N.