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In "DeltaA B C","B C"=1,sin"A"/2="x"1,si...

In `"DeltaA B C","B C"=1,sin"A"/2="x"_1,sin"B"/2="x"_2cos"A"/2="x"_3` and `cosB/2=x_4w i t h(x ?1/(x_2))^(2007)-((x_3)/(x_4))^(2006)=0` Length of side AC is equal to: 1/2 (b) 1 (c) 2 (d) cant be determined If `/_A=90^0,t h e na r e aof A B C` is 1/2 sq. units (b) 1/3 sq. units 1 sq.. units (d) 2 sq. units

A

`1/2`

B

1

C

2

D

can't be determined

Text Solution

Verified by Experts

The correct Answer is:
B
In given `DeltaABC" both " A/2and B/2"lie strictly in "(0,pi/2)and sin x " always increasing in "(0,pi/2)` whereas cos x is always decreasing in `(0,pi/2)`.
So, if `A/2ltB/2`
`rArr sin(A)/2gt sin(B)/2`
`or x_1gtx_2`
`and x_3ltx_4`
`or 1/x^3gt1/x^4`
So, `x_1^2007x_4^2006=x_2^2007x_3^2006` is not valid.
Similarly for `A/2ltB/2`
`rArr sin(A)/2sin(B)/2`
`rArr x_1ltx_2`
`and 1/x^3lt1/x^4`
For this also `x_1^2007x_4^2006=x_2^2007x_3^2006` is not valid.
So,`(x_1/x_2)^2007-(x_3/x_4)^2006=0" is possible only when " A/2=B/2`.
`rArr x_1=x_2and 1/x_3=1/x_4`
Hence, `DeltaABC` is isosceles with `angleABC=angleCAB`.
`rArr BC=AC=1"unit"`
if `angleA=90^@`
Area, `A=1/2BCxxAC=1/2" sq. units"`
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