please solve the question by showing its explanation
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Consider ∆BDA
angle ADB+DBA+BAD=180
angle BAD=26
angle DAC=90-26=64
consider∆EAC
angle DEC is an exterior angle
therefore
angle DEC=angleACE+angleEAC
angle EAC=angle DAC
angle DEC-angle ACE=64
angle ADB+DBA+BAD=180
angle BAD=26
angle DAC=90-26=64
consider∆EAC
angle DEC is an exterior angle
therefore
angle DEC=angleACE+angleEAC
angle EAC=angle DAC
angle DEC-angle ACE=64
Thatsomeone:
thanks for selecting my answer as brainliest
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