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With usual notion, if in triangle A B C ...

With usual notion, if in triangle `A B C ,` `(b+c)/(11)=(c+a)/(12)=(a+b)/(13),t h e np rov et h a t` `(cosA)/7=(cosB)/(19)=(cosC)/(25)`

A

`7:12:95`

B

`7:19:25`

C

`7:91:52`

D

`71:92:5`

Text Solution

Verified by Experts

The correct Answer is:
B
Let `(b+c)/(11)=(c+a)/(12)=(a+b)/(13)=k`
`rArrb+c=11k,c+a=12k,a+b=13k` …(i)
and `2(a+b+c)36k`
On solving Eqs . (i) and (ii) , we get
`a=7k,b=6k,c=5k`
Now , `cosA=(b^(2)+c^(2)-a^(2))/(2bc)`
`=(36k^(2)+25k^(2)-49k^(2))/(60k^(2))=(1)/(5)`
`cosB=(a^(2)+c^(2)-b^(2))/(2ac)`
`=(49k^(2)+25k^(2)-36k^(2))/(70k^(2))=(38)/(70)=(19)/(35)`
and `cosC=(a^(2)+b^(2)-c^(2))/(2ab)=(49k^(2)+36k^(2)-25k^(2))/(84k^(2))`
`=(60)/(84)=(5)/(7)`
`thereforecosA:cosBcosC=(1)/(5):(19)/(35):(5)/(7)=7:19:25`
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