In the diagram, ABCD is a circle O. The quadratic OBCD is a rhombus such that \(\angle\)ADO = \(\angle\)OBA = y and \(\angle\)BAD = t. Find;
(a) the value of t;
(b) the value of y;
(c) \(\angle\)ADC.
Attached Figure / Diagram
Verified Correct Answer: .
Step-by-Step Mathematical Solution
(a) to find t
\(\angle\)BCD = \(\angle\)BOD = 2t (angle at the centre = twice that at the circumference)
\(\angle\)BCD = \(\angle\)BAD = 180º (opposite angles of a cyclic quad. are supplementary)
2t + t = 180º
3t = 180º
t = 60º
(b) to find the value of y
\(\angle\)BOD = 2t = 2 x 60 = 120º
\(\angle\)BOD = \(\angle\)ODB (equal radii)
⇒ \(\angle\)OBD +\(\angle\)ODB + 2t = 180º ( sum of
Source: Myschool Past Questions Archive (Record #71921)