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In triangle $\triangle ABC$, a point $D$ is on $\overline{AC}$ so that $AB = AD$ and $\angle ABC - \angle ACB = 32^{\circ}$. Find $\angle DBC$ in degrees.

 

 

[asy] size(125); pair C = (0,0); pair B = (3,0); pair A = intersectionpoint (C--C+100dir(35),B--B+100dir(115)); pair D = 0.65*C+0.35*A; draw(A--B--C--cycle); draw(B--D); dot(Label("$A$", A, N)); dot(Label("$B$", B, SE)); dot(Label("$C$", C, SW)); dot(Label("$D$", D, NW)); [/asy]

 Sep 22, 2019
edited by Guest  Sep 22, 2019
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In triangle \(\triangle ABC\), a point \(D\) is on \(\overline{AC}\) so that \(AB = AD\) and \(\angle ABC - \angle ACB = 32^{\circ}\).
Find \(\angle DBC\) in degrees.

 

\(\begin{array}{|rcll|} \hline \angle ABD &=& \angle ADB \quad & | \quad \angle ADB = \angle DBC + \angle ACB \\ \angle ABD &=& \angle DBC + \angle ACB \quad & | \quad \angle ABD = \angle ABC - \angle DBC \\ \angle ABC - \angle DBC &=& \angle DBC + \angle ACB \\ 2\angle DBC &=& \angle ABC - \angle ACB \quad & | \quad \angle ABC - \angle ACB = 32^{\circ} \\ 2\angle DBC &=& 32^{\circ} \\ \mathbf{\angle DBC} &=& \mathbf{16^{\circ}} \\ \hline \end{array} \)

 

laugh

 Sep 23, 2019

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