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Given (b+c)/(11)=(c+a)/(12)=(a+b)/(13) f...

Given `(b+c)/(11)=(c+a)/(12)=(a+b)/(13)` for a `DeltaABC` with usual notation. If `(cos A)/(alpha)=(cos B)/(beta)=(cos C)/(gamma)` then the ordered traiad, `(alpha,beta,gamma)` has a vlaue

A

(19,7,25)

B

(3,4,5)

C

(5,12,13)

D

(7,19,25)

Text Solution

Verified by Experts

The correct Answer is:
D
Given `(b+c)/(11)=(c+a)/(12)=(a+b)/(13)= lambdaa` (say)

`b+c=11 lambda, c+a=12lambda and a+b=13lambda " "...(ii)`
`rArr 2(a+b+c)=36lambda`
`a+b+c=18lambda " "...(ii)`
From Eqs (i) and (ii), we get
`a=7lambda,b=6lambda,c=5lambda`
Now,
`cosA=(b^(2)+c^(2)-a^(2))/(2ac)=(lambda^(2)[36+25-49])/(60lambda^(2))=(12)/(60)=(1)/(5)`
`cosB=(b^(2)+c^(2)-a^(2))/(2ac)=(lambda^(2)[49+25-36])/(70lambda^(2))=(19)/(35)`
`and cosC=(b^(2)+c^(2)-a^(2))/(2ac)=(lambda^(2)[49+25-36])/(84lambda^(2))`
`=(60)/(84)=(5)/(7)`
Thus, `cos A=(1)/(5)=(7)/(35)`
`cosB=(19)/(35),cosC=(25)/(35)`
` (cosA)/(7)=(cosB)/(19)=(cosC)/(25)=(1)/(35)`
`rArr(alpha, beta, gamma)=(7,19,25)`
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