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In a triangle ABC, if cos A cos B + sin ...

In a triangle `ABC`, if `cos A cos B + sin A sin B sin C = 1`, then `a:b:c` is equal to

Text Solution

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Given `cos A cos+B sin A sin B sin 1`
`rArr sin C=(1-cos A cosB)/(sin A sin B)" " ....(i)`
`rArr (1-cos A cosB)/(sin A sin B)le" "[ :. le cos (A-B)]`
`rArr 1-cos A cos B le sin A sin B " " 1 le cos (A-B)`
`rArr cos (A-B)ge1`
`rArr cos(A-B=1)" " [ :. as cos (theta) le1]`
`rArr A-B=0`
On putting A=B Eq. (i) we get
` sin C=(1-cos^(2)B)/(sin^(2)A)`
`rArr sin C=1`
`rArr C=pi//2`
Now, `A+B+C=pi`
`rArr A+B=(pi)/(2)rArr A=(pi)/(4)" " [ :. A=B C =(pi)/(2)]`
`:. sin A: sin B: sinC: = "sin"(pi)/(4): "sin"(pi)/(4): "sin"(pi)/(2)`
`rArr a:b:c:(1)/(sqrt(2)):(1)/(sqrt(2)):1`
` =1 :1:sqrt(2)`
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