Construct a triangle 𝐴𝐵𝐶 with side 𝐵𝐶 = 7𝑐𝑚,∠𝐵 = 45°, ∠𝐴 = 105°.Then construct another triangle whose sides are $\frac{3}{4}$ times of the corresponding sides of the  △𝐴𝐵𝐶.

First draw a rough diagram to find angle 𝐶

In ∆𝐴𝐵𝐶 

⇒ ∠𝐴 + ∠𝐵 + ∠𝐶 = 180° 

⇒ 105° + 45° + ∠𝐶 = 180° 

⇒ ∠𝐶 = 180° − 150° 

⇒ ∠𝐶 = 30° 

Steps to draw ∆𝐴𝐵𝐶 

1. Draw the base of side 7 𝑐𝑚 

2. Draw ∠𝐵 = 45° 

3. 𝐷𝑟𝑎𝑤 ∠𝐶 = 30° 

4. At point 𝐴 the two rays intersect

Now we need to make triangle which is $\frac{3}{4}$ times its size

Steps of construction:

1. Draw any ray 𝐵𝑋 making an acute angle with 𝐵𝐶 on side opposite to vertex A

2. Make 4 points 𝐵1 , 𝐵2 , 𝐵3 , 𝐵4 on BC so that 𝐵𝐵1 = 𝐵𝐵2 = 𝐵𝐵3 = 𝐵𝐵4.

3. Join 𝐵 and draw a line through parallel to to intersect extend At .

4. Draw a line through 𝐶' parallel to 𝐴𝐶 to intersect 𝐴𝐵 extended at 𝐴'.

Hence ∆𝐴'𝐵𝐶' is the required triangle.