Answer:
\[BC=7\text{ }cm,\,\angle B=45{}^\circ ,\text{ }\angle \text{ }A=105{}^\circ \] \[\angle C=180{}^\circ -(\angle B+\angle A)\] \[=180{}^\circ -(45{}^\circ +105{}^\circ )\] \[=180{}^\circ -150{}^\circ \] \[=30{}^\circ \] Steps of construction- (i) Draw a line segment \[BC=7\text{ }cm\]. (ii) Draw an angle \[45{}^\circ \] at B and \[30{}^\circ \] at C. They intersect at A. (iii) Draw an acute angle at B. (iv) Divide angle ray in 4 equal parts as \[{{B}_{1}},{{B}_{2}},{{B}_{3}}\] and \[{{B}_{4}}\]. (v) Join \[{{B}_{4}}\] to C?. (vi) From \[{{B}_{3}}\]draw a line parallel to \[{{B}_{4}}C\] intersecting BC at C?. (vii) Draw another line parallel to CA from C' intersecting AB ray at A'. (viii) \[\Delta \text{ }A'BC'\] is required triangle such that \[\Delta \text{ }A'BC'\sim \Delta \,ABC\] with \[A'B=\frac{3}{4}AB\].
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