Draw triangle 𝐴𝐵𝐶 such that 𝐵𝐶 = 5 𝑐𝑚, ∠𝐴𝐵𝐶 = 60°.∠𝐴𝐶𝐵 = 30º. Now construct Δ𝐴'𝐵𝐶' corresponds to Δ𝐴𝐵𝐶 with 𝐴'𝐵 : 𝐴𝐵 = 3 : 2

1. Draw a line segment BC  of length 5 cm. 

2. Draw the angles of 60°  and 30⁰  on the points B  and C  respectively which intersect each other at A.  

3. ΔABC is the given triangle. 

4. Draw a ray BX  making an acute angle with BC. 

5. Locate three points B₁, B₂ and B₃ on line segment BX. Such that BB₁ = B₁B₂ = B₂B₃

6. Join B₂C. 

7. Draw B₃C' || B₂C to intersect the extended line BC' at C'.

8. Through C draw a line parallel to AC intersecting extended line segment BA at A'.

ΔA'BC' is the required triangle.