किसी `triangleABC` में यदि `(b+c)/(11) = (c+a)/(12) = (a+b)/(13)` तो सिद्ध करे कि `(cosA)/(7) = (cosB)/(19) = (cosC)/(25)`

Explanation

मानाकि `(b+c)/(11) = (c+a)/(12) = (a+c)/(13) = k`
तो `b +c = 11k, c +a = 12k, a +b = 13k,` इन सबो को जोड़ने पर ,
`2(a + b +c) = 36k therefore a + b +c = 18k therefore a = 7k, b = 6k, c = 5k`
अब `(cosA)/(7) = (b^(2) + c^(2)-a^(2))/(2bc.7) = (36k^(2) + 25k^(2) - 49k^(2))/(2.6k.5.k7) = (1)/(35)`
`(cosB)/(19) = (c^(2) + a^(2) -b^(2))/(2ca.19) = (25k^(2) + 49k^(2) - 36k^(2))/(2*5k.7k.19) = (1)/(35)`
इसी प्रकार ` = (cosC)/(25) = (1)/(35) therefore (cosA)/(7) = (cosB)/(19) = (cosC)/(25)`