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In `DeltaABC,BC=1,sin.(A)/2=x_1,sin.(B)/2=x_2,cos.(A)/2=x_3andcos.(B)/2=x_4" with "(x_1/x_2)^2007-(x_3/x_4)^2006=0` If `angleA=90^@`, then area of `DeltaABC` isA. 1/2 sq. unitsB. 1/3 sq. unitsC. 1 sq. unitsD. 2sq. Units |
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Answer» Correct Answer - A 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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