Given:-

∠BEA = 32°

∠BEC = 27° and ∠EDB = 105°

To find:-

a) ∠BAE, b) ∠CBA and c) ∠CBE

Solution:-

a)

∠EDB = 1/2 arc(BAE) ....(inscribed angle)

105 = 1/2 arc(BAE)

105 × 2 = arc(BAE)

210° = arc(BAE) ...... (1)

now,

arc(BCDE) + arc(BAE) = 360° ....(full circle)

arc(BCDE) + 210 = 360

arc(BCDE) = 360 - 210

arc(BCDE) = 150° ......(2)

now,

∠BAE = 1/2 arc(BCDE) ....(inscribed angle)

∠BAE = 1/2 × 150

∠BAE = 75° .....(3)

b)

∠BEA = 1/2 arc(AB)

32° = 1/2 arc(AB)

32 × 2 = arc(AB)

64° = arc(AB) .....(4)

And,

∠BEC = 1/2 × arc(BC) ....(inscribed angle)

27 = 1/2 arc(BC)

54° = arc(BC) ......(5)

now,

arc(AB) + arc(BC) + arc(AEDC) = 360° ...(full circle)

arc(ABC) + arc(AEDC) = 360°

64 + 54 + arc(AEDC) = 360

118 + arc(AEDC) = 360

arc(AEDC) = 360 - 118

arc(AEDC) = 242° .....(6)

now, ∠CBA = 1/2 arc(AEDC)

∠CBA = 1/2 × 242

∠CBA = 121° .......(7)

c)

arc(BC) + arc(BAE) + arc(CDE) = 360° ....(from 1 and 5)

54 + 210 + arc(CDE) = 360

264 + arc(CDE) = 360

arc(CDE) = 360 - 264

arc(CDE) = 96° .....(8)

now,

∠CBE = 1/2 arc(CDE) ....inscribed angle

∠CBE = 1/2 × 96

∠CBE = 48° ...... (9)

• ANSWER

a) ∠BAE = 75°, b) ∠ CBA = 121° and c) ∠CBE = 48°