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avatar+965 

Thanks!

 Jul 20, 2020
 #1
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0

Angle DBC = angle DBA = x, angle ABC = 2x

AB = AC, so angle BCD = 2x

Angle ADB external to triangle BDC, so angle ADB = 3x

BD = BC, so angle BDC = 90 - x

angle ADB + angle BDC = 180 ==> 3x + (90 - x) = 180 ==> x = 45

==> A = 90 - x = \(\boxed{45}\)

 Jul 20, 2020
 #2
avatar+25600 
+1

Please help ASAP

 

\(\begin{array}{|rcll|} \hline AB=AC \\ \hline \angle ABC &=& \angle ACB \quad | \quad \angle ABC =2x \\ 2x &=& \angle ACB \\ \mathbf{\angle ACB} &=& \mathbf{2x} \\ \hline \end{array} \begin{array}{|rcll|} \hline BD=BC \\ \hline \angle BDC &=& \angle ACB \quad | \quad \angle ACB =2x \\ \mathbf{\angle BDC} &=& \mathbf{2x} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \angle ADB &=& 180^\circ - \angle BDC \quad | \quad \mathbf{\angle BDC=2x} \\ \mathbf{\angle ADB} &=& \mathbf{180^\circ - 2x} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{ \text{In $\triangle$ BCD }: } \\ \hline 2x+2x+x &=& 180^\circ \\ 5x &=& 180^\circ \\\\ x &=& \dfrac{180^\circ}{5} \\\\ \mathbf{x} &=& \mathbf{36^\circ} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{ \text{In $\triangle$ ADB }: } \\ \hline A+ (180^\circ-2x) +x &=& 180^\circ \\ A+ 180^\circ-2x +x &=& 180^\circ \\ A+ 180^\circ-x &=& 180^\circ \\ A -x &=& 0 \\ A &=& x \quad | \quad \mathbf{x=36^\circ} \\ \mathbf{A} &=& \mathbf{36^\circ} \\ \hline \end{array} \)

 

laugh

 Jul 21, 2020

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