Question

This is the last question i have of this! i have been stuck on it all of yesterday

Answer 1

Given:

`\bar{AD}\text{ is a diameter of }\odot B`

Therefore,

`m\angle A=(1)/(2)m\angle CBD`

Thus

`\begin{gathered} \text{If m}\angle A=35^0, \\ m\angle CBD=2* m\angle A=2*35^0=70^0 \\ \therefore m\angle CBD=70^0 \end{gathered}`

Also,

`\begin{gathered} \text{If m}\angle CBD=100^0, \\ \therefore m\angle A=(1)/(2)*100^0=50^0 \\ \therefore m\angle A=50^0 \end{gathered}`

Finally,

`\begin{gathered} \text{If m}\angle A=x, \\ m\angle CBD=2* x=2x \\ \therefore m\angle CBD=2x \end{gathered}`

Final answers

`\begin{gathered} a)\text{ }m\angle CBD=70^0 \\ b)\text{ }m\angle A=50^0 \\ c)\text{ }m\angle CBD=2x \end{gathered}`