In the adjoining figure DE is internal bisector of $∠$ D, find the measure of $∠$ x.

155^{\circ}

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105^{\circ}

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115^{\circ}

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130^{\circ}

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Solution

The correct option is A $155^{\circ}$
$∠ A C B + ∠ C B A + ∠ B A C = 180 °$
$∠ A C B = 180 ° - 70 ° - 35 ° = 75 °$
$∠ E C D = 180 ° - ∠ A C B = 180 ° - 75 ° = 105 °$
$∠ G F D = ∠ G C D + ∠ C D F$ (Exterior angle property)
$∠ G F D = 105 ° + 50 ° = 155 °$

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Similar questions

∠ D

∠ x

DE is internal bisector of angle D,find the measure of angle x

∠ A

∠ B

O

∠ C = 100^{\circ}

∠ D = 60^{\circ}

∠ A O B

In the given figure, O is the centre of a circle and $∠ A O B = 130^{\circ} . T h e n, ∠ A C B = ?$

(a) $50^{\circ}$

(b) $65^{\circ}$

(d) $155^{\circ}$

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