The sides BC, CA and AB of $Δ A B C$ are produced in order to form exterior angles $∠ A C D$ , $∠ B A F$ and $∠ C B E$ respectively, then $∠ B A F + ∠ A C D + ∠ C B E$ is

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Solution

The correct option is C

By angle sum property,
$x + y + z = 180^{\circ}$
$∠ B A F + ∠ A C D + ∠ C B E$
$= (180^{\circ} - x) + (180^{\circ} - y) + (180^{\circ} - z)$
$= 540^{\circ} - (x + y + z) = 540^{\circ} - 180^{\circ} = 360^{\circ}$

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The sides BC, CA and AB of ΔABC are produced in order to form exterior angles ∠ACD, ∠BAF and ∠CBE respectively, then ∠BAF+∠ACD+∠CBE is

The correct option is C By angle sum property, x+y+z=180∘ ∠BAF+∠ACD+∠CBE =(180∘−x)+(180∘−y)+(180∘−z) =540∘−(x+y+z)=540∘−180∘=360∘

The sides BC, CA and AB of $Δ A B C$ are produced in order to form exterior angles $∠ A C D$ , $∠ B A F$ and $∠ C B E$ respectively, then $∠ B A F + ∠ A C D + ∠ C B E$ is

The correct option is C

By angle sum property,
$x + y + z = 180^{\circ}$
$∠ B A F + ∠ A C D + ∠ C B E$
$= (180^{\circ} - x) + (180^{\circ} - y) + (180^{\circ} - z)$
$= 540^{\circ} - (x + y + z) = 540^{\circ} - 180^{\circ} = 360^{\circ}$