Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.

We need to find ∠C:

Provided that:

∠B = 45°, ∠A = 105°

As we know, the sum of all interior angles in a triangle is 180 degrees.

∠A+∠B +∠C = 180°

105°+45°+∠C = 180°

∠C = 180° − 150°

∠C = 30°

As a result, from the property of triangle, we get ∠C = 30°

Construction Procedure:

The required triangle can be constructed as follows:

1. Draw a triangle ABC with base BC = 7 cm, ∠B = 45°, and ∠C = 30°.

2. Draw a ray BX on the opposite side of vertex A that makes an acute angle with BC.

3. As 4 is greater in 4 and 3, so now locate 4 points on the ray BX, such as B₁, B₂, B₃, B₄.

4. Now join the points B₃C.

5. Draw a line parallel to B₃C through B₄ that intersects the extended line BC at C’.

6. Draw a line parallel to the line AC through C’, intersecting the extended line segment at C’.

7. As a result, ΔA’BC’ is the required triangle.

Justification:

It is possible to justify the construction of the given problem by proving that

Since 4/3 is the scale factor, we need to prove

A’B = (4/3)AB

BC’ = (4/3)BC

A’C’= (4/3)AC

We get A’C’ || AC from the construction

Now in ΔA’BC’ and ΔABC,

Therefore, ∠A’C’B = ∠ACB as these are corresponding angles

∠B = ∠B as it is the common angle

Therefore, from the AA similarity criterion ΔA’BC’ ∼ ΔABC (From AA similarity criterion)

Since the similar triangle’s corresponding sides are in the same ratio, therefore it becomes

 A’B/AB = BC’/BC= A’C’/AC

Hence, it becomes A’B/AB = BC’/BC= A’C’/AC = 4/3

As a result, the above construction is justified.