Question

Help solve this math problem

Answer 1

Check the picture below.

so let's do this one a bit oddly kinda.

we know AB is a diameter, that means the arc made by AB is simply 180°, so the arcs AE + ED + DB = 180°.

By the inscribed angle theorem, as you see there, the arcDB is simply 2ɑ and by the same theorem ED is 88°, so the arc AE is just 180° - ED - DB, let's keep in mind that AE is the "far arc" whilst DB is the "near arc", and we know the external angle they make is 18°.

`18~~ = ~~\cfrac{(\stackrel{ far~arc }{180-88-2\alpha})~~ - ~~\stackrel{ near~arc }{2\alpha}}{2} \implies 36=(180-88-2\alpha)-2\alpha \\\\\\ 36=92-4\alpha\implies 36+4\alpha=92\implies 4\alpha=56\implies \alpha=\cfrac{56}{4}\implies \boxed{\alpha=14}`