Question

I would like some help thank you :)

Answer 1

`\angle AE = 32^(\circ)`

`\angle EAD = 212^(\circ)`

`\angle BE = 133^(\circ)`

`\angle BCE = 227^(\circ)`

`\angle AED = 180^(\circ)`

`\angle BD = 79^(\circ)`

Explanation:

The central angle of a circle is equal to 360º, whose formula in this case is:

`\angle AB + \angle BC + \angle CD + \angle DE + \angle EA = 360^(\circ)`

In addition, the following conditions are known from figure:

`\angle BC = 47^(\circ)`,
`\angle DE = 148^(\circ)`

`\angle DE + \angle EA = 180^(\circ)`

`\angle CD + \angle DE = 180^(\circ)`

`\angle AB + \angle BC + \angle CD = 180^(\circ)`

Now, the system of equations is now solved:

`\angle EA = 180^(\circ)-\angle DE`

`\angle EA = 180^(\circ)-148^(\circ)`

`\angle EA = 32^(\circ)`

`\angle CD = 180^(\circ)-\angle DE`

`\angle CD = 180^(\circ)-148^(\circ)`

`\angle CD = 32^(\circ)`

`\angle AB = 180^(\circ) - \angle BC - \angle CD`

`\angle AB = 180^(\circ)-47^(\circ)-32^(\circ)`

`\angle AB = 101^(\circ)`

The answers are described herein:

`\angle AE = 32^(\circ)`

`\angle EAD = 212^(\circ)`

`\angle BE = 133^(\circ)`

`\angle BCE = 227^(\circ)`

`\angle AED = 180^(\circ)`

`\angle BD = 79^(\circ)`