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If the area of a triangle is given Delta...

If the area of a triangle is given `Delta` and angle C is given and if the value of the side c opposite to angle C is minimum then

A

`a=sqrt((2Delta)/(sin C))`

B

`b=sqrt((2Delta)/(sin C))`

C

`a=(4Delta)/(sin C)`

D

`b=(4Delta)/(sin^(2)C)`

Text Solution

Verified by Experts

The correct Answer is:
A, B
`c^(2)=a^(2)+b^(2)-2ab cos C`
`=(a-b)^(2)+2ab (1-cos C)`
`=(a-b)^(2)+(4 Delta)/(sin C)(1-cos C)`
Hence, for c to minimum, a = b
Also `Delta =(1)/(2)ab sin C rArr a^(2)=(2Delta)/(sin C)=b^(2)`
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