6. Draw a triangle ABC with side BC = 7 cm, B = 45°, A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of D ABC.
Steps of construction :-
(i) Draw a line segment BC.
(ii) Now draw an angle B = 45o and C = 30o and draw rays in these directions.
(iii) Name the intersection of these lines as A.
(iv) Thus is the required triangle.
(v) Draw a line BX which makes an acute angle with BC and is opposite of vertex A.
(vi) Cut four equal parts of line BX namely BB1, BB2, BB3, BB4.
(vii) Now join B3 to C. Draw a line B4C' parallel to B3C.
(viii) And then draw a line B'C' parallel to BC.
Hence AB'C' is the required triangle.