Question

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

Answer 1

$Stepsofconstruction:$
$1.Constructatriangle$ $ABC$ $having$$BC=7cm,\angle A={45}^o$ and $\angle B={105}^o$.
$2.Draw$$Bx$ $rayat$ $B$ $toformacuteangle.$
$3.Markpoints$ $B_1,B_2,B_3,B_4$ $from$ $B$. $Draw$ $B_4{C}^{\prime }$.
$4.Draw$ $B_{4}C||B_3{C}^{\prime }$. $Draw$ $AC||A^{\prime }{C}^{\prime }$.
$5.Now,$ $\Delta A^{\prime }B{C}^{\prime }|||\Delta ABC$ $isconstructed.$
$\Delta A^{\prime }B{C}^{\prime }:\Delta ABC$