In fig AC is diameter ∠ACB = 400 Find ∠ BDC
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Step-by-step explanation:
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Answer:
Given: ∠BAD=65
o
,∠ABD=70
o
,∠BDC=45
o
(i) ∵ ABCD is a cyclic quadrilateral.
In Δ ABD,
∠ BDA +∠ DAB +∠ ABD=180
o
By using sum property of Δ
s
∴ ∠BDA =180
o
−(65
o
+70
o
)
=180
o
−135
o
=45
o
Now from Δ ACD,
∠ADC =∠ADB+∠BDC
=45
o
+45
o
(∵∠ BDA=∠ADB=45
o
)
Hence, ∠D makes right angle belongs in semi-cricle therefore AC is a diameter of the circle.
(ii) ∠ACB=∠ADB (Angles in the same segment of a circle)
∴ ∠ACB=45
o
(∵∠ ADB=45
o
).
Step-by-step explanation:
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