Question

In the diagram given alongside, AC is the diameter of the circle, with centre O. CD and BE are parallel. ∠ AOB = 80° and ∠ ACE = 10°. Calculate: (i) ∠ BEC (ii) ∠ BCD (iii) ∠ CED.

Answer 1

From the figure, we have
∠ AOB = 80°
∠ ACE = 10°

(i) ∠ BOC = 180° - ∠ AOB
⇒ ∠ BOC = 180° -  80°
⇒ ∠ BOC = 100°

∠ BEC = `1/2`∠ BOC    ...( ∵ ∠ Subtended at the centre and ∠ subtend by E by arc BC)
⇒ ∠ BEC = `1/2` x 100°
⇒ ∠ BEC = 50°

(ii) ∠ ACB = `1/2` ∠ AOB     ...( ∵ ∠s Subtended by arc AB at the centre at C)
⇒ ∠ ACB = `1/2 xx 80°` 
⇒ ∠ ACB = 40°

∠ ECD = ∠ BEC     ...( ∵ Alternate ∠s as CD || BE)
∠ ECD = 50°

⇒ ∠ BCD = ∠ ACB + ∠ ECA + ∠ ECD 
⇒ ∠ BCD = 40° + 10° + 50°
⇒ ∠ BCD = 100°

(iii) BCDE is a cyclic quadrilateral,     ....(∵ Its opposite ∠s are supplementary)
⇒ ∠ BED = 180° - ∠ BCD     
⇒ ∠ BED = 180° - 100°       ....( From ii)
⇒ ∠ BED = 80°

⇒ ∠ BEC + ∠ CED = 80°
⇒  ∠ CED = 80° - ∠ BED
⇒  ∠ CED = 80° - 50°
⇒  ∠ CED = 30°            ....( From i)