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Find X.
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Let triangle be ABC.
Right angled at B.
AB = X,
BC = Y + 100,
tan theta = opposite side/adjacent side.
In TRiangle ABD,
tan 45 = y/x
=) 1 = y/x
=) x = y - - 1)
In TRiangle ABC,
tan 60 = (y+100)/x
=) √3 = (y+100)/x
=) √3x = y+100
=) √3x - 100 = y - - 2)
Comparing both eqs,
=) x = √3x - 100
=) 100 = x(√3 - 1)
=) 100/(√3 - 1) * ((√3 +1) /√3 +1) =x
=) 100(√3 +1) / 2 = x
=) 50(√3 +1) m = X.
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