Draw ABC in which side BC = 7 cm, B = 45°, A = 105°. Then construct another triangle whose sides are 3/5 of the corresponding sides of ABC.

Given: ∠B = 45° and ∠A = 105°

We know that sum of all interior angles in a triangle is 180°.

∴ ∠A + ∠B + ∠C = 180°

⇒ 105° + 45° + ∠C = 180°

⇒ ∠C = 180° - 150°

⇒ ∠C = 30°

Construction Steps:

Step 1: Draw a ΔABC with side BC = 7cm, ∠B = 45° and ∠C = 30°.

Step 2: Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

Step 3: Locate 5 (the greater of 3 and 5 in 3/5) points B₁, B₂, B₃, B₄ and B₅ on BX so that BB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B₅.

Step 4: Join B₅C and draw a line through B₃ (the 3rd point, 3 being smaller of 3 and 5 in 3/5) parallel to B₅C to intersect BC at C′.

Step 5: Draw a line through C′ parallel to the line CA to intersect BA at A′.

Then, ΔA′BC′ is the required triangle.

Justification:

After dividing the line segment BC in the given ratio, BC’/C’C = 3/2

⇒ BC/BC’ = BC’ + C’C / BC’ = 1 + C’C/BC’ = 1 + 2/3 = 5/3

∴ BC/BC’ = 5/3

Also, C’A’ is parallel to CA.

∴ ΔA’BC’ ~ ΔABC

So, A’B/AB = A’C’/AC = BC’/BC = 3/5

This justifies the construction.