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Q47. Given that $\angle$ACD is 4^° than $\angle$CDE and $\angle$CDE is 22^° more than $\angle$DAC. Also, the sum of $\angle$BCE , and $\angle$BDE is 26^° and $\angle$EBF = 80°. Find the value of x.
         

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$$\begin{gathered} Given: \\ \angle ACD=4°+\angle CDE.........\left(i\right) \\ \angle CDE=22°+\angle DAC............\left(ii\right) \\ \angle BCE+\angle BDE=26°...............\left(iii\right) \\ \angle EBF=80° \\ Now, \\ In ∆ACD \\ \angle ACD+\angle CDA+\angle DAC=180° \\ From equation \left(i\right) and \left(ii\right) \\ 4°+22°+\angle DAC+22°+\angle DAC+\angle DAC=180° \left[\angle CDA=\angle CDE\right] \\ 48°+3\angle DAC=180° \\ \angle DAC=44° \\ From e{q}^n\left(ii\right) \\ \angle CDE=66° \\ From e{q}^n \left(i\right) \\ \angle ACD=70° \\ Now, \\ \angle CDE=\angle BDE+\angle BDC=66°.......\left(iv\right) \\ \angle ACD=\angle BCE+\angle ECD=70°..........\left(v\right) \\ Now adding e{q}^n \left(iv\right)\& \left(v\right) \\ \angle BDE+\angle BDC+\angle BDE+\angle BDC=136° \\ 26°+\angle ECD+\angle BDC=136° \left[\angle BCE+\angle BDE=26°\right] \\ \angle ECD+\angle BDC=110° \\ Now in ∆FCD \\ \angle FCD+\angle FDC+\angle CFD=180° \\ \angle CFD=180°-110° \\ \angle CFD=70° \\ \angle CFD=\angle BFE=70 ° \left[Vertically Opposite Angles\right] \\ Now in ∆BEF \\ \angle BFE+x+\angle EBF=180° \\ 70°+x+80°=180° \\ x=180°-\left(150°\right) \\ \boxed{\mathbf{x}\mathbf{=}\mathbf{30}\mathbf{°}} Answer \end{gathered}$$

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