Respuesta :

caylus
Hello,

If we draw the parallele of AB passing through B
∠SBT=x+x=2x
∠BCT=x ==>∠BCD=180°-x

∠SBT+∠BCD=2x+180°-x=180°+x


dalendrk
[tex]Let's\ \angle TCB=y^o\\\\\angle SBT\ is\ the\ exterior\ angle\ of\ \Delta TCB,\ therefore\ |\angle SBT|=x^o+y^o.\\\\|\angle\ BCD|\ +\ y^o=180^o-angles\ on\ one\ side\ of\ a\ straight\ line\\\\|\angle BCD|=180^o-y^o\\\\therefor\\\\\angle SBT+\angle BCD=x^o+y^o+180^o-y^o=x^o+180^o\\\\\boxed{Answer:\angle SBT+\angle BCD=x^o+180^o}[/tex]

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