∠BAD+∠BCD=180° nbsp; nbsp;[Sum nbsp;of nbsp;the nbsp;opposite nbsp;angles nbsp;of nbsp;the nbsp;cyclic nbsp;quadrilateral nbsp;is nbsp;1 ...

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Byju's Answer

Standard X

Mathematics

Angles in Alternate Segment

Find ∠CDE[2 m...

Question

Find $∠ C D E$
[2 marks]

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Solution

$∠ B A D + ∠ B C D = 180 ° [S u m o f t h e o p p o s i t e a n g l e s o f t h e c y c l i c q u a d r i l a t e r a l i s 180 d e g r e e s .] ∠ B A D + 110 ° = 180 ° ∠ B A D = 70 °$
[1 mark]

$∠ C D E = ∠ B A D = 70 ° [T a n g e n t s e c a n t s e g m e n t t h e o r e m]$
[1 mark]

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Find ∠CDE [2 marks]

∠BAD+∠BCD=180° [Sum of the opposite angles of the cyclic quadrilateral is 180 degrees.]∠BAD+110°=180°∠BAD=70° [1 mark] ∠CDE=∠BAD=70°[Tangent secant segment theorem] [1 mark]

Find $∠ C D E$
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