Verified answer

[tex]The ΔA′BC′ whose sides are 43 of the corresponding sides of ΔABC can be drawn as follows:

Step 1: Draw a ΔABC with side BC=6cm,AB=5cm,∠ABC=60∘

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

Step 3: Locate 4 points, B1,B2,B3,B4 on line segment BX.

Step 4: Join B4C and draw a line

through B3, parallel to B4C intersecting BC at C′.

Step 5: Draw a line through C′ parallel to AC intersecting AB at A′.

The triangle A′BC′ is the required triangle.

\[/tex]

Step-by-step explanation:

Steps of construction :

Draw a line segment PQ of length 5 cm.

At P, draw a ray PX making 35° with PQ.

At Q, draw a ray QY making 105° with PQ.

Rays PX and QY will intersect at point R.

Triangle PQR is now constructed.