Here,

According to the question,

In ΔPQR,

Using angle sum property of triangle,

∠RPQ = 180 – (∠PQR + ∠QRP)

= 180 – (105 + 40)

= 180 – 145

∠RPQ = 35°

We have to draw figure using following steps of construction:

Step 1: Draw a line segment PQ of 5 cm

Step 2: Now, from point P draw a ray PX making an angle of 35° from PQ.

Step 3: From Q, draw a ray QY from PQ making 105° angle. and intersecting PX at R

Step 4: The ray QY will intersect with PX at point R. Hence ΔABC is the required triangle.