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.