8) Draw ZABC = 120° Draw its angle bisector.
Answers
Step-by-step explanation:
Steps of Construction:
Steps of Construction:1.Construct a line segment BC.
Steps of Construction:1.Construct a line segment BC.2.Taking B as centre and some suitable radius construct an arc which meets BC at the point P.
Steps of Construction:1.Construct a line segment BC.2.Taking B as centre and some suitable radius construct an arc which meets BC at the point P.3. Taking P as centre and with same radius cut off the arcs PQ and QR.
Steps of Construction:1.Construct a line segment BC.2.Taking B as centre and some suitable radius construct an arc which meets BC at the point P.3. Taking P as centre and with same radius cut off the arcs PQ and QR.4.Now join BR and produce it to point A
Steps of Construction:1.Construct a line segment BC.2.Taking B as centre and some suitable radius construct an arc which meets BC at the point P.3. Taking P as centre and with same radius cut off the arcs PQ and QR.4.Now join BR and produce it to point A∠ABC=120
Steps of Construction:1.Construct a line segment BC.2.Taking B as centre and some suitable radius construct an arc which meets BC at the point P.3. Taking P as centre and with same radius cut off the arcs PQ and QR.4.Now join BR and produce it to point A∠ABC=120 o
Steps of Construction:1.Construct a line segment BC.2.Taking B as centre and some suitable radius construct an arc which meets BC at the point P.3. Taking P as centre and with same radius cut off the arcs PQ and QR.4.Now join BR and produce it to point A∠ABC=120 o
Steps of Construction:1.Construct a line segment BC.2.Taking B as centre and some suitable radius construct an arc which meets BC at the point P.3. Taking P as centre and with same radius cut off the arcs PQ and QR.4.Now join BR and produce it to point A∠ABC=120 o 5.Taking P and R as centres construct two arcs which intersect each other at the points S.
Steps of Construction:1.Construct a line segment BC.2.Taking B as centre and some suitable radius construct an arc which meets BC at the point P.3. Taking P as centre and with same radius cut off the arcs PQ and QR.4.Now join BR and produce it to point A∠ABC=120 o 5.Taking P and R as centres construct two arcs which intersect each other at the points S.6.Now join BS and produce it to point D.
Steps of Construction:1.Construct a line segment BC.2.Taking B as centre and some suitable radius construct an arc which meets BC at the point P.3. Taking P as centre and with same radius cut off the arcs PQ and QR.4.Now join BR and produce it to point A∠ABC=120 o 5.Taking P and R as centres construct two arcs which intersect each other at the points S.6.Now join BS and produce it to point D.Here BD is the bisector of ∠ABC
Steps of Construction:1.Construct a line segment BC.2.Taking B as centre and some suitable radius construct an arc which meets BC at the point P.3. Taking P as centre and with same radius cut off the arcs PQ and QR.4.Now join BR and produce it to point A∠ABC=120 o 5.Taking P and R as centres construct two arcs which intersect each other at the points S.6.Now join BS and produce it to point D.Here BD is the bisector of ∠ABCBy measuring each angle we get to know that is it 60
Steps of Construction:1.Construct a line segment BC.2.Taking B as centre and some suitable radius construct an arc which meets BC at the point P.3. Taking P as centre and with same radius cut off the arcs PQ and QR.4.Now join BR and produce it to point A∠ABC=120 o 5.Taking P and R as centres construct two arcs which intersect each other at the points S.6.Now join BS and produce it to point D.Here BD is the bisector of ∠ABCBy measuring each angle we get to know that is it 60 o
Steps of Construction:1.Construct a line segment BC.2.Taking B as centre and some suitable radius construct an arc which meets BC at the point P.3. Taking P as centre and with same radius cut off the arcs PQ and QR.4.Now join BR and produce it to point A∠ABC=120 o 5.Taking P and R as centres construct two arcs which intersect each other at the points S.6.Now join BS and produce it to point D.Here BD is the bisector of ∠ABCBy measuring each angle we get to know that is it 60 o yes,both the angles are of equal measure.
