Draw a triangle ABC with side BC = 6 cm, ∠C = 30° and ∠A = 105°. Then construct another triangle whose sides are 2/3 times the corresponding sides of ΔABC.

Given: BC = 6 cm, ∠C = 30° and ∠A = 105°
We know that in ΔABC, ∠A + ∠B + ∠C = 180°
Therefore, ∠B = 180° − (∠A + ∠C) = 180° − (105° + 30°)
=> 45°
Steps of construction:

1. Draw a line BC = 6 cm.
2. Draw a ray CN making an angle of 30° at C.
3. Draw a ray BM making an angle of 45° at B.
4. Locate the point of intersection of rays CN and BM and name it as A.
5. ABC is the triangle whose similar triangle is to be drawn.
6. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
7. Locate 3 (Greater of 2 and 3 in 2/3) points B1, B2 and B3 on BX so that BB1 = B1B2 = B2B3.
8. Join B3C and draw a line through B2 (Smaller of 2 and 3 in 2/3) parallel to B3C to intersect BC at C′.
9. Draw a line through C′ parallel to the line CA to intersect BA at A′.
10. A′BC′ is the required similar triangle whose sides are 2/3 times the corresponding sides of ΔABC.