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0
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Hello, can you guys find BA please?

May 8, 2021

#1
0

CA and DB are diameters of the circle and EC, ED, EB, and EA are radii which means EC=ED=EB=EA which means they are isosceles triangles and since angle CED is congruent to angle AEB we have triangle AEB is congruent to triangle CED.

Now we need ti find z.

Let's focus on triangle CED. Since triangle CED is an isosceles triangle, angle C and D are congruent which means

16z+12+2z+4+2z+4 = 180

20z+20=180

20z=160

z=8

Then solve for angle D

2z+4 = D

16+4 = D

D = 20

And since triangles CED and AEB are congruent Angle D = Angle A so Angle A = 20

:)))))))))))))))))))))

Wait oops you asked for mBA

If thats the case it is quite simple. We already know z = 8 and angle CED = angle AEB and since CED is 140 that means AEB is 140

SO IN CONCLUSION

mBA = 140

Wow i wrote alote

May 8, 2021
edited by Guest  May 8, 2021
#2
+32859
+1

(2z+4)*2       +       16z+12 = 180       (inscribed angles intercept arcs twice their measure)

z =  8

Arc BA = 16z+12 = 140 degrees

May 8, 2021
#4
0

Arc CB = 2(2z + 4)

Arc CB + arc CD = 180º

(16z + 12) + 2(2z + 4) = 180

z = 8

Arc CD = arc AB = 16 * 8 + 12 = 140º

May 8, 2021