Because I couldn't get any help on the last problem I got a whole bunch more. Could someone please help with this?
Let AB be a diameter of a circle centered at O. Let E be a point on the circle, and let the tangent at B intersect the tangent at E and AE at C and D, respectively. If angle BAE = 43 degrees, find angle CED, in degrees.
Angle BAE = 43° ...so minor arc BE will = 86°
And angle CBE = 1/2 of this arc = 43°
But.......since CB and CE are tangents to the circle from a common point C....they are equal
Thus...in triangle BCE....BC = EC...and the angles opposite these sides are also equal...
So angle CBE = angle BEC
And note that in triangle AEB.....angle AEB intercepts a diameter....so its measure = 90°
But angle AEB + angle BEC + angle CED = 180°.....so....
90 + 43 + angle CED = 180
133 + angle CED = 180 subtract 133 from each side
angle CED = 47°