question_answer

Given below are the steps of construction a triangle ABC with side \[BC=6\text{ }cm,\] \[\angle B={{60}^{o}}\], \[\angle A={{150}^{o}}\]and a triangle whose sides are (3/2) times the corresponding ; sides of A ABC. Which of the following steps of construction is INCORRECT?   Steps of Construction

Step I: Draw BC = 6 cm.

Step II: At B construct \[\angle CBX={{60}^{o}}\] and at C construct

Suppose BX and CY intersect at A. \[\Delta \,ABC\]so obtained is the given triangle.

Step III: Construct an obtuse angle \[\angle CBZ\] at B on opposite side of vertex A of \[\Delta \,ABC.\]

Step IV: Mark-off three (greater 3 of 2 in 3/2) points \[{{B}_{1}},{{B}_{2}},{{B}_{3}},\] on BZ such that\[B{{B}_{1}}={{B}_{1}}{{B}_{2}}={{B}_{2}}{{B}_{3}}.\].

Step V: Join \[{{B}_{2}}\] (the second point) to C and draw a line through \[{{B}_{3}}\] parallel to \[{{B}_{2}}C,\] intersecting the extended line segment BC at C'.

Step VI: Draw a line through C' parallel to CA intersecting the extended line segment BA at A'.

Triangle A'B'C so obtained is the required triangle such that \[\frac{A'B}{AB}=\frac{BC'}{BC}=\frac{A'C'}{AC}=\frac{3}{2}\]

A)  Step III           

B)  Step IV                       

C)  Step V         

D)         Step II