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# help plz

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In triangle $ABC$, $\angle ABC = 2 \alpha$ and $\angle ACB = 140 ^\circ - \alpha$. $D$ is a point on line segment $AB$ such that $CB = BD$. What is the measure (in degrees) of $\angle DCA$?

Jan 14, 2021

#1
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In triangle $$ABC$$, $$\angle ABC = 2 \alpha$$ and $$\angle ACB = 140 ^\circ - \alpha$$.
$$D$$ is a point on line segment $$AB$$ such that $$CB = BD$$.
What is the measure (in degrees) of $$\angle DCA$$?

$$\text{Let \angle DCA=x} \\ \text{Let \angle BDC=\angle BCD= \beta}$$

$$\begin{array}{|rcll|} \hline \mathbf{2\alpha + 2\beta} &=& \mathbf{180^\circ} \\ \alpha + \beta &=& 90^\circ \\ \mathbf{\beta} &=& \mathbf{90^\circ -\alpha} \\ \hline \end{array}$$

$$\begin{array}{|rcll|} \hline \mathbf{140^\circ-\alpha} &=& \mathbf{x + \beta} \quad | \quad \mathbf{\beta=90^\circ -\alpha} \\ 140^\circ-\alpha &=& x + 90^\circ -\alpha \\ 140^\circ&=& x + 90^\circ \\ x &=& 140^\circ-90^\circ \\ \mathbf{x} &=& \mathbf{50^\circ} \\ \hline \end{array}$$

Jan 15, 2021