A triangle if its perimeter is 10.4 cm and two angles are 45° and 120°.

Step 1:

Draw a line XY=10.4 i.e. the perimeter.

Step 2:

Construct an angle equal to ∠B=45° and another angle equal to ∠C=120°

Step 3:

Bisect these angles and name the intersecting point as A.

Step 4:

Construct perpendicular bisectors of AX and AY and name the PQ and RS respectively.

Step 5:

Name the intersecting point of PQ and XY as B and RS the intersecting point of RS and XY as C.

Join AB and AC.

ABC is the required triangle.

Justification

As B is on line PQ which is the perpendicular bisector of AX,

AB+BC+CA = XB+BC+CY=XY

Then,∠BAX=∠AXB (as in triangle AXB, AB is equal to XB)

As ∠ABC is the external angle of triangle AXB

Then,∠ABC=∠BAX + ∠AXB(exterior angle sum property)

∠ABC=∠AXB+∠AXB

∠ABC=2∠AXB=45° or ∠B.

Similarly,

∠ACB=2∠CAY=120° or ∠C.

Thus the construction is justified.