Question

gprove that :angle FAB+angle ABC+angle BCD+angle CDE+angle DEF+angle EFA=720 degrees ​

Answer 1

Answer:

i gave you the next question answer as well as this question answer.

Answer 2

Answer:

$Sum \: of \: \angle {FAB}\\ + \angle {ABC} + \angle {BCD}\\ + \angle {CDE} + \angle {DEF}\\ + \angle {EFA} \\=(\angle {FAE} + \angle {EAD} + \angle {DAC} +\angle {CAB} ) \\+ \angle {ABC} \\+ (\angle {BCA} + \angle {ACD})\\ +(\angle {CDA} + \angle {ADE} )\\ + (\angle {DEA} + \angle {AEF} )\\+ \angle {EFA}$

$= ( \angle 1 +\angle 2 +\angle 3 +\angle 4 )\\+ \angle 5 + ( \angle 6 +\angle 7 ) \\+ (\angle 8 +\angle 9 ) \\+ (\angle 10 +\angle 11 )\\+\angle 12$

/* Rearranging the angles , we get

$= ( \angle 12 +\angle 1 +\angle 11 )\\+(\angle 10 +\angle 2+\angle 9 )\\+ (\angle 3 +\angle 7 +\angle 9 )\\+(\angle 4 +\angle 5 +\angle 6)\\= Sum \: of \:the \: angles \: in \: triangles ( \triangle FAE +</p><p>\triangle EAD +\triangle DAC +\triangle CAB )\\= 180\degree +180\degree+180\degree+180\degree\\= 4\times 180\degree$

$\boxed { \pink { Sum \: of \: the \: angles \: in \: a \: triangle = 180\degree }}$

$= 720\degree$

$Hence \: proved .$

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