If in a triangle of base 'a', the ratio of the other two sides is r ( <1).Show that the altitude of the triangle is less than or equal to `(ar)/(1-r^2)`

Explanation

Given that in `Delta ABC`, base `= a and (c)/(b) = r`. We have to find the altitude `h`.

We have, in `Delta ABD`
`h = c sin B = (c a sin B)/(a)`
`= (c sin A sin B)/(sin (B + C))`
`= (c sin A sin B sin (B -C))/(sin (B + C) sin (B - C))`
`= (c sin A sin B sin (B - C))/(sin^(2) B - sin^(2)C)`
`=(c(a)/(k) (b)/(k) sin (B -C))/((b^(2))/(k^(2)) - (c^(2))/(k^(2)))`
`= (abc sin (B - C))/(b^(2) - c^(2))`
`= (a((c)/(b)) sin (B - C))/(1 - ((c)/(b))^(2))`
`= (ar sin (B - C))/(1 - r^(2))`
`le (ar)/(1 - r^(2)) " " [ :' sin (B - C) le 1]`
`:. h le (ar)/(1 - r^(2))`