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# Draw a triangle ABC with side BC = 7 cm, $$\angle$$ B=45°, $$\angle$$ A=105°;. Then construct a triangle whose sides are $$\frac{4}3$$ times the corresponding sides of $$\triangle$$ ABC.

Steps to construct the figure by given data:
Step 1: Draw a line segment BC=7 cm.

Step 2: Draw $$\angle$$ ABC=45° ; and $$\angle ACB=30°$$ ; ,i.e, $$\angle BAC=105°$$ ;

Step 3: We get $$\triangle$$ ABC.

Step 4: Now, draw a ray BX making an acute angle with BC.

Step 5: Mark 4 points $$B_1,B_2,B_3$$ and $$B_4$$ on the ray BX so that $$BB_1=B_1B_2=B_2B_3=B_3B_4$$ and Join $$B_3C$$.

Step 6: Through $$B_4$$, draw a line $$B_4C’$$ parallel to $$B_3C$$, intersecting the extended line segment BC at C’.

Step 7: Through C’, draw a line A’C’ parallel to CA, intersecting the extended line segment BA at A’.

Thus, ΔA’BC’, is the required triangle.

JUSTIFICATION:

In $$\triangle A’BC’ and \triangle$$ ABC,

$$\angle ABC=\angle A’BC’$$(common)

$$\angle ACB= \angle A’C’B$$ (corresponding angles)

$$\therefore \triangle ABC ~ A’BC’$$ (By AA similarity)

$$\therefore \frac{AB}{A’B}=\frac{AC}{A’C’}=\frac{BC}{BC’}$$

But, $$\frac{BC}{BC’}=\frac{BB_3}{BB_4}=\frac{3}4 ∴ \frac{BC’}{BC}=\frac{4}3$$

$$⇒ \frac{A’B}{AB}=\frac{A’C’}{AC}=\frac{BC’}{BC}=\frac{4}3 .$$

$$\therefore$$ Sides of the new triangle formed are $$\frac{4}{3}$$ times the corresponding sides of the first triangle.