To find BC we use the pythagorean theorem:
[tex]\begin{gathered} BC^2=CA^2-BA^2=5^2-4^2=25-16=9 \\ BC=3 \end{gathered}[/tex]
Now, notice that
[tex]\begin{gathered} CA=CD+DA=BC+DA \\ DA=CA-BC=5-3 \\ DA=2 \end{gathered}[/tex]
To find BCD we use the trigonometric function sin
[tex]\begin{gathered} \sin (\measuredangle\text{BCD)}=\text{ sin(}\measuredangle\text{BCA)}=\frac{BA}{CA}=\frac{4}{5} \\ \measuredangle BCD=sin^{-1}(\frac{4}{5})=53.13^{\circ} \end{gathered}[/tex]
For the last question we use the definition of the arc with a given angle
[tex]\begin{gathered} \text{mBD =BC}\cdot\measuredangle BCD\frac{2\pi}{360}=3\cdot53.13\frac{\pi}{180} \\ \text{mBD}=2.78 \end{gathered}[/tex]
