Question

please answer this question in detail.​

Answer 1

Step-by-step explanation:

Given: ∠BAD=65o,∠ABD=70o,∠BDC=45o

(i) ∵ ABCD is a cyclic quadrilateral.

In Δ ABD,

∠ BDA +∠ DAB +∠ ABD=180o By using sum property of Δs

∴ ∠BDA =180o−(65o+70o)

=180o−135o

=45o

Now from Δ ACD,

∠ADC =∠ADB+∠BDC

=45o+45o     (∵∠ BDA=∠ADB=45o)

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=45o   (∵∠ ADB=45o

Answer 2

Answer:

HOPE THIS HELPS.....

Step-by-step explanation:

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