AOB is the diameter of a circle. Chords AC and BD are produced to meet at E. If angle COD = 40 degree, then find angle CED - 20/50/70/90 degrees.
Answers
∠CED= 70° if AC & DB are two chords and meet at E , AB is Diameter & ∠COD = 40°
Step-by-step explanation:
Lets join OC & OD
now in Δ AOC
OA = OC ( Radius)
=> ∠CAO = ∠ACO = x
Simialrly
∠DBO = ∠BDO = y ( as OD = OB = Radius)
ABCD is a cylic Quadrilateral
=> ∠BAC + ∠BDC = 180° ( oposite angles)
∠BAC = ∠OAC as O lies on AB ( as AB is diameter)
∠BDC = ∠BDO + ∠COD
=> ∠OAC + ∠BDO + ∠COD = 180°
=> x + y + ∠COD = 180°
=> ∠COD = 180° - x - y
∠COD = ∠CDO as ( OC = OD = Radius)
=> ∠COD = 180° - x - y
in ΔOCD
∠DCO + ∠COD + ∠DOC= 180°
=> 2 ( 180° - x - y) + 40° = 180°
=> 180° - x - y = 70°
in ΔAEB
∠EAB + ∠EBA + ∠AEB = 180°
∠AEB = ∠CED
=> x + y + ∠CED = 180°
=>∠CED = 180° - x - y
=> ∠CED = 70°
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Answer:
LUTION
Since ACE is a straight line, we have
ZACB+2BCE = 180°
90+2BCE=180°
[ ZACB is in a semicircle]
ZBCE = 90°.
Also, 2DBC = COD=(×40) = 20
[angle at centre = 2x angle at a point on a circle]
ZEBC=2DBC=20".
Now, in AEBC, we have
ZEBC+2BCE+4CEB= 180°
⇒ 20 +90 + <CED = 180°
ZCED=180-110¹ = 70°.
[ ZCEB=LCED]
Hence, CED=70".