Question

Point B and C lie on tangent to the circle drawn at point A . Chord AD = Chord ED .
If m (arc EPF) =1/2 m (arc AQD) and m (arc DRE) = 84 ; then determine.
1 ) DAC 2 ) FDA 3) FED 4 ) BAF​

Answer 1

Point B and C lie on tangent to the circle drawn at point A . Chord AD = Chord ED .

If m (arc EPF) =1/2 m (arc AQD) and m (arc DRE) = 84 ; then determine.

1 ) DAC 2 ) FDA 3) FED 4 ) BAF​

• Construction: Join OD, OE, OA, OF, FD and AF.
• 1 ) DAC
• In △EOD and △AOD, we have:
• AO=EO       (Radii of the circle)
• OD=OD       (Common)
• ED=AD        (Given)
• ∴△EOD ≅ △AOD      (By SSS congruency)
• i.e., ∠DEO=∠DAO   (CPCT)
• Also, ∠DOE=∠AOD=84°  (CPCT, m(arcDRE) = 84°)
• Now, in△EOD, we have:
• ∠EOD+∠ODE+∠OED=180°  (Angle sum property)
• ⇒2∠OED=180°−84°    
• ⇒∠OED=48°
• Hence, ∠OED=∠DAO=48°
• ∠EAC=90°  (Radius is perpendicular to tangent)
• ∠DAC=∠EAC−∠DAO=90°−48°=42°
• ∠EOF=12∠AOD=42°
• In △EOD and △AOD, we have:
• AO=EO       (Radii of the circle)
• OD=OD       (Common)
• ED=AD        (Given)
• ∴△EOD ≅ △AOD      (By SSS congruency)
• i.e., ∠DEO=∠DAO   (CPCT)
• Also, ∠DOE=∠AOD=84°  (CPCT, m(arcDRE) = 84°)
• Now, in△EOD, we have:
• ∠EOD+∠ODE+∠OED=180°  (Angle sum property)
• ⇒2∠OED=180°-84°    
• ⇒∠OED=48°
• Hence, ∠OED=∠DAO=48°
• ∠EAC=90°  (Radius is perpendicular to tangent)
• ∠DAC=∠EAC-∠DAO=90°-48°=42°
• ∠EOF=12
• ∠AOD=42°
• 2 ) FDA
• ∠EOF+∠EOD+∠DOA+∠AOF=360°  (Complete angle)
• ⇒∠AOF=360°− ∠EOF−∠EOD−∠DOA
• ⇒∠AOF=360°−42°−84°−84°=150°
• ∠EOF+∠EOD+∠DOA+∠AOF=360°  (Complete angle)
• ⇒∠AOF=360°- ∠EOF-∠EOD-∠DOA
• ⇒∠AOF=360°-42°-84°-84°=150°
• We know that the angle subtended by an arc at the centre is twice the angle subtended at any part of the circle.
• i.e., ∠FDA=12∠AFO=75°∠FDA=12∠AFO=75°
• 3) FED
• In△EOF, we have:
• ∠EOF+∠FEO+∠EFO=180°  (Angle sum property)
• ⇒2∠FEO=180°−42°    
• ⇒∠FEO=69°
• So, ∠FED=∠OED+∠FEO=48°+69°=117°
• In△EOF, we have:
• ∠EOF+∠FEO+∠EFO=180°  (Angle sum property)
• ⇒2∠FEO=180°-42°    
• ⇒∠FEO=69°
• So, ∠FED=∠OED+∠FEO=48°+69°=117°
• 4 ) BAF​
• In△AOF, we have:
• ∠AOF+∠FAO+∠OFA=180°  (Angle sum property)
• ⇒2∠OAF=180°−150°    
• ⇒∠OED=15°
• So, ∠OAB=90°  (Radius is perpendicular to tangent)∴∠FAB=∠OAB−∠OAF=90°−15°=75°

Answer 2

Answer:

Answer in photo

Step-by-step explanation:

Please check photo