Question

[**not drawn to scale**]What is the length of radius BC?What is the length of segment DA?What is the measure of BCD?What is the measure of mBD =?

Answer 1

To find BC we use the pythagorean theorem:

`\begin{gathered} BC^2=CA^2-BA^2=5^2-4^2=25-16=9 \\ BC=3 \end{gathered}`

Now, notice that

`\begin{gathered} CA=CD+DA=BC+DA \\ DA=CA-BC=5-3 \\ DA=2 \end{gathered}`

To find BCD we use the trigonometric function sin

`\begin{gathered} \sin (\measuredangle\text{BCD)}=\text{ sin(}\measuredangle\text{BCA)}=(BA)/(CA)=(4)/(5) \\ \measuredangle BCD=sin^(-1)((4)/(5))=53.13^(\circ) \end{gathered}`

For the last question we use the definition of the arc with a given angle

`\begin{gathered} \text{mBD =BC}\cdot\measuredangle BCD(2\pi)/(360)=3\cdot53.13(\pi)/(180) \\ \text{mBD}=2.78 \end{gathered}`