Question

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construct a triangle PQR with PQ =3.6 cm QR=3.2cm and ∆Q=120°​

Answer 1

Answer:

take a sacle and compas

with the help of scale draw a line PQ 3.6 CM and then draw a line 3.2 CM(with the help of protector make a dot of 120 degree from the line of PQ)

Draw the line QR 3.2 CM

Answer 2

Answer:

Δ $PQR$ is the required triangle.

Step-by-step explanation:

Draw rough sketch of Δ $PQR.$

Step $1:$ Draw a line segment $PQ=3.6cm.$

Step $2:$ Draw a ray $QX$ such that ∠ $Q=120$°

Step $3:$ With $Q$ as centre and radius $3.2cm$ draw an arc to intersect $QX$ at $R.$

Step $4:$ Join $RP.$

Thus, Δ $PQR$ is the required triangle.