Question

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

Answer 1

Answer:

Step-by-step explanation:

To divide a line segment AB in the ratio m:n, a ray AX making an acute ∠BAX is drawn and then (m+n) points are marked at equal distances on ray AX

m=3 and n=5

∴Minimum number of points to be marked on AX

⇒m+n=3+5=8