Question : Given below is a circle with O. Find the value of ∠BCD in terms of a and b.
(a) 180°+(a-b/2)
(b) 180°-(a+b/2)
(c) 180°-(a-b/2)
(d) None of these.
Doubt by Vikrant
Solution :
∠1+b=180° (Linear Pair)
∠1=180°-b — (1)
In ΔOAD
In ΔOAD
OA=OD (Radii of the same circle)
∠2=∠3 (Angles opposite to equal sides of a triangle are equal)
Also,
∠2=∠3 (Angles opposite to equal sides of a triangle are equal)
Also,
∠1+∠2+∠3=180° (ASP)
∠1+∠3+∠3=180° [∵∠2=∠3]
∠1+2∠3=180°
2∠3=180°-∠1
∠1+2∠3=180°
2∠3=180°-∠1
2∠3=180°-(180°-b) [From equation (1)]
2∠3=180°-180°+b
2∠3=b
∠3=b/2 — (2)
2∠3=180°-180°+b
2∠3=b
∠3=b/2 — (2)
In ΔADB
∠3+∠4=a [Exterior Angle Property of the Triangle]
b/2+∠4=a [Using equation (2)]
∠4=a-b/2 — (3)
∠4=a-b/2 — (3)
Now,
ABCD is a cyclic Quadrilateral
∠BCD+∠4=180° [Sum of opposite angles of a cyclic quadrilateral is supplementary]∠BCD=180°-∠4
∠BCD=180°-(a-b/2) [Using equation (3)]
Hence, c) 180°-(a-b/2), would be the correct option.