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किसी त्रिभुज ABC में यदि `(b+c)/(11)=(c+a)/(12)=(a+b)/(13)` तो सिद्ध कीजिए कि `(cos A)/(7)=(cosB)/(19)=(cosC)/(25)`

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माना `(b+c)/(11)=(c+a)/(12)=(a+b)/(13)=k` (माना)
`rArr b+c=11k, c+a=12k, a+b=13k`
सभी को जोड़ने पर
`2(a+b+c)=36krArr a+b+c=18k`
इसलिए `a=7k,b=6k` तथा `c=5k`
अब क्योंकि `cos A=(b^2+c^2-a^2)/(2bc)=(36k^2-49k^2)/(2.6k.5k)=(1)/(5)`
`rArr (cosA)/(7)=(1)/(5xx7)=(1)/(35)`
इसी प्रकार `(cosB)/(19)=(1)/(35)` तथा `(cosC)/(25)=(1)/(35)`
`rArr (cosA)/(7)=(cosB)/(19)=(cosC)/(25)`

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