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secondary 3 | E Maths
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Can someone help me with part d
Date Posted: 3 years ago
Views: 170
∠BAD = ∠BDA = 65°
(base angles of isosceles ∆, since AB = BD)
∠BCD + ∠BAD = 180°
(opposite angles of a cyclic quadrilateral add up to 180°)
Or
(angles in opposite segments are supplementary)
∠BCD = 180° - 65° = 115°
(base angles of isosceles ∆, since AB = BD)
∠BCD + ∠BAD = 180°
(opposite angles of a cyclic quadrilateral add up to 180°)
Or
(angles in opposite segments are supplementary)
∠BCD = 180° - 65° = 115°
See 2 Answers
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Date Posted:
3 years ago
Sorry, i found angle ACD... please add back 65.. answer 115°
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Date Posted:
3 years ago
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