Discussion
O is a point on line AB. Rays OC and OD are drawn on the same side of AB such that \(OC \perp OD\), \(\angle AOC = x\) and \(\angle BOD = 3x\). Then, \(\angle BOD\) is equal to :
Answer: A
\(\angle BOD\) is \(67\frac{1}{2}^\circ\).
Angles AOC, COD and DOB lie on the straight line AB, so they add up to 180°, and \(\angle COD = 90^\circ\) because OC is perpendicular to OD:
\[ \begin{aligned} x + 90^\circ + 3x &= 180^\circ \\ 4x &= 90^\circ \\ x &= 22\tfrac{1}{2}^\circ \end{aligned} \]
So \(\angle BOD = 3x = 67\frac{1}{2}^\circ\). The option \(22\frac{1}{2}^\circ\) is x itself, \(\angle AOC\), not \(\angle BOD\).
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