Line AD bisecting ΔCAB results in two triangles. The first triangle ΔADB has angle measurements m_ADB = 100° and
m B = 48°. The total angle of a triangle is 180 degrees. Hence, the measurement of angle A on the triangle ΔADB is
[tex]\begin{gathered} \angle A+\angle B+\angle ADB=180 \\ \angle A=180-\angle B-\angle ADB \\ \angle A=180-100-48=32 \end{gathered}[/tex]The triangle ΔCAD has no given angles on it. But we can find first the angle of m CDA by knowing that angle CDA and angle ADB as supplementary angles.
[tex]\begin{gathered} \angle CDA+\angle ADB=180 \\ \angle CDA=180-\angle ADB \\ \angle CDA=180-100=80 \end{gathered}[/tex]Using sine law, the angles DAB, CAD, BDA, and CDA are related as
[tex]\frac{\angle DAB}{\angle BDA}=\frac{\angle CAD}{\angle CDA}[/tex]Solve for the value of angle CAD
[tex]\angle CAD=\frac{32}{100}\times80=25.6[/tex]Hence, angle CAD has an angle measurement of 25.6°.
