to divide a line segment BC internally in the ratio 3:5 , we draw a ray BX such that angle CBX is an acute angle . What will be t

to divide a line segment BC internally in the ratio 3:5 , we draw a ray BX such that angle CBX is an acute angle . What will be the minimum number of points to be located at equal distances , on ray BX??

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2 thoughts on “to divide a line segment BC internally in the ratio 3:5 , we draw a ray BX such that angle CBX is an acute angle . What will be t”

  1. Correct option is

    C

    8

    The minimum number of points=3+5=8

    In this process, once line AX is drawn, it is divided into 8 equal parts using a pair of compass.

    The points are marked from point A towards X. The last point is then joined to point B to line XB.

    Lines are then drawn parallel to XB and passing through the points that were marked on AX. These lines can be drawn using set squares to ensure they are parallel.

    These parallel lines will divide line AB into 8 equal parts. So, to divide the line in the ratio 3:5, the first three portions will be taken and last 5 left.

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