In `Delta ABC` if `r_1=2r_2=3r_3` and `D` is the mid point of `BC` then `Cos/_ADC` is (a) `7/25` (b) `-7/25` (c) `24/25` (d) `-24/25`

Options

A. `(7)/(25)`
B. `-(7)/(25)`
C. `(24)/(25)`
D. `-(24)/(25)`

Answer

D

Explanation

`r_(1)=2r_(1)=3r_(3)`
`rArr (Delta)/(s-a)=(2Delta)/(s-b)=(3Delta)/(s-c)=(Delta)/(k)` (say)
`rArr s-a=k, s-b=2k, s-c=3k`
`rArr 3s-(a+b+c)=6k rArr s=6k`
`rArr (a)/(5)=(b)/(4)=(c )/(3)=k`
So, `a^(2)=b^(2)+c^(2)`
`Delta ABC` is right angled `Delta, angle A =90^(@)` and D is midpoint of BC,
AD = DC (radius of circumicrcle)
`rArr angle DAC = C rArr angle ADC = 180^(@)-2C`
`rArr cos angle ADC`
`=cos(180^(@)-2C)`
`=-cos 2C`
`-(2cos^(2)C-1)`
`=1-2cos^(2)C`
`=1-2xx(16)/(25)=(-7)/(25)`
`["from " Delta ABC, cos C=(b)/(a)=(4)/(5)]`