What is the value of x, if ∆ADE - ∆ACB, /_DEC = 105° and /_ECB = 65°?
a. 45°
b. 60°
c. 13°
d. 40°
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Answer:
45 degree
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).
hope it helps mark as brainliest pls
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