Question

Given: Circle PB52°РMAD =mBD =mBAC =:: 52°.: 90°:: 128°:: 142°.: 232°:: 308°

Answer 1

From the circle given, it can be observed that AC is the diameter of the circle and it divides the circle into two equal parts. The total angle in a semi-circle is 180°. It then follows that

`arcAD+arcDC=arcAC`
`\begin{gathered} \text{note that} \\ arcAC=180^0(\text{angle of a semicircle)} \\ arcDC=90^0(\text{given)} \end{gathered}`
`\begin{gathered} \text{Therefore,} \\ arcAD+arcDC=arcAC \\ arcAD+90^0=180^0 \\ arcAD=180^0-90^0 \\ arcAD=90^0 \end{gathered}`
`\begin{gathered} \text{From the circle, it can be seen that:} \\ arcBD=arcBA+arcAD \\ \text{note that } \\ arcBA=52^0(\text{given)} \\ arcAD=90^0(\text{calculated earlier)} \end{gathered}`
`\begin{gathered} \text{Therefore,} \\ arcBD=52^0+90^0 \\ arcBD=142^0 \end{gathered}`
`\begin{gathered} \text{From the given circle, it can be seen that} \\ arcBA+arcAD+arcDC=arc\text{BAC} \end{gathered}`
`\begin{gathered} \text{Therefore,} \\ 52^0+90^0+90^0=\text{arcBAC} \\ 232^0=\text{arcBAC} \end{gathered}`

Hence, arcAD = 90°, arc BD = 142°, and arc BAC = 232°