find angle DEC and DCE
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in BDC
BDC + DCB + DBC= 180
100 +25 + DBC = 180
DBC = 55
<A+ <B+ <C = 180
55 + 55 + C= 180
C= 70
DCE + DCB = 70
25 + DCE = 70
DCE = 45
CDE = 25( alternate angle)
in triangle CDE
CDE + DCE + CED = 180
25 + 45 + ced = 180
CED = 110
BDC + DCB + DBC= 180
100 +25 + DBC = 180
DBC = 55
<A+ <B+ <C = 180
55 + 55 + C= 180
C= 70
DCE + DCB = 70
25 + DCE = 70
DCE = 45
CDE = 25( alternate angle)
in triangle CDE
CDE + DCE + CED = 180
25 + 45 + ced = 180
CED = 110
netra45:
thank you :)
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Hope it will help ...
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thanks alot :)
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