Mathematics
In the figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25°. Find :
(i) ∠CAD
(ii) ∠CBD
(iii) ∠ADC
Circles
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Answer
(i) We know that,
Exterior angle of a cyclic quadrilateral is equal to interior opposite angle.
∠BAD = Exterior ∠BCE = 80°.
From figure,
∠CAD = ∠BAD - ∠BAC = 80° - 25° = 55°.
Hence, ∠CAD = 55°.
(ii) We know that,
Angles in same segment are equal.
∴ ∠CBD = ∠CAD = 55°.
Hence, ∠CBD = 55°.
(iii) Since, AB ∥ DC,
∠ACD = ∠BAC = 25° [Alternate angles are equal]
In triangle ADC,
By angle sum property of triangle,
⇒ ∠ADC + ∠CAD + ∠ACD = 180°
⇒ ∠ADC + 55° + 25° = 180°
⇒ ∠ADC + 80° = 180°
⇒ ∠ADC = 180° - 80°
⇒ ∠ADC = 100°.
Hence, ∠ADC = 100°.
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