**Ex 11.1, 6**

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 Δ ABC.

In Δ ABC

∠B = 45°, ∠A = 105°

Sum of all interior angles in a triangle is 180°.

∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°

∠C = 180° − 150°

∠C = 30°

Steps of construction

- Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.
- Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
- Locate 4 points (as 4 is greater in 4 and 3), B_1, B_2, B_3, B_4, on BX.
- Join B_3C. Draw a line through B_4 parallel to B_3C intersecting extended BC at C'.
- Through C', draw a line parallel to AC intersecting extended line segment at C'.

ΔA'BC' is the required triangle.

**Justification**

Here,

(BC^′)/BC=(BB_4)/(BB_3 ) = 4/3

Also, A’C’ is parallel to AC

So, the will make the same angle with line BC

∴ ∠ A’C’B = ∠ ACB

Now,

In Δ A’BC’ and ABC

∠ B = ∠ B

∠ A’C’B = ∠ ACB

**Δ**** A’BC’ ∼ ****Δ**** ABC**

Since corresponding sides of

similar triangles are in the same ratio

(A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC

So, **(A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC** **=4/3****.**

This justifies the construction.