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In the sides of a triangle are in the ratio `1:sqrt(3):2,` then the measure of its greatest angle is `pi/6` (b)`pi/3` (c) `pi/2` (d) `(2pi)/3`

A

`pi/6`

B

`pi/2`

C

`pi/3`

D

`(2pi)/(3)`

Text Solution

Verified by Experts

The correct Answer is:
B

Let `a=1x,b=sqrt(3)x,c=2x`
now,by cosine law,
`2cb cos A=c^2+b^2-a^2`
=`2xx(2x)xx(sqrt(3)x) cos A=(2x)^2+(sqrt(3)x)^2-(x)^2`
`=>cosA=(4+3-1)/(2xx2xxsqrt(3))`
`=>cos A=(sqrt(3))/2`
`=>A=30^@`

Solving for B
...
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