Answer:
\[\angle B=45{}^\circ \] and \[\angle A=105{}^\circ \] \[\because \] Sum of angles of triangle is \[180{}^\circ \] \[\therefore \angle A+\angle B+\angle C=180{}^\circ \] \[105{}^\circ +45{}^\circ +\angle C=180{}^\circ \] \[\Rightarrow \angle C=180{}^\circ -(105{}^\circ +45{}^\circ )\] \[\Rightarrow \angle C=30{}^\circ \] Steps of construction: (i) Draw a line segment \[BC=7\text{ }cm\] (ii) Construct \[\angle B=45{}^\circ \] and \[\angle C=30{}^\circ \] (iii) A is the intersecting point of ray through B and C. Thus, \[\Delta \text{ }ABC\] is obtained. (iv) Draw D on BC such that \[BD=\frac{4}{5}BC=\left( \frac{4}{5}\times 7 \right)cm=\frac{28}{5}cm=5.6\,cm\] Then, \[\Delta \text{ }BDE\] is the required triangle similar to \[\Delta \text{ }ABC\]such that each side of \[\Delta \text{ }BDE\] is \[\frac{4}{5}\] times the corresponding side of\[\Delta \text{ }ABC\].
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