Question

$\angle AOF={120}^o,\angle BOF={60}^o$ and $\angle BOD={90}^o$

Then find the unknown angles in the figure, where AB,CD and EF are straight lines intersecting at O.

• A. $\angle COF={90}^o,\angle COA={90}^o,\angle AOE={60}^o,\angle EOD={60}^o$
• B. $\angle COF={30}^o,\angle COA={90}^o,\angle AOE={60}^o,\angle EOD={30}^o$
• C. $\angle COF={60}^o,\angle COA={60}^o,\angle AOE={90}^o,\angle EOD={90}^o$
• D. $\angle COF={30}^o,\angle COA={30}^o,\angle AOE={60}^o,\angle EOD={30}^o$

Answer 1

The correct option is B $\angle COF={30}^o,\angle COA={90}^o,\angle AOE={60}^o,\angle EOD={30}^o$

$\angle AOF={120}^o,\angle BOF={60}^o,\angle BOD={90}^o$ ....... [Given]
$\angle BOD=\angle COA$ (vertically opposite angles)
$\therefore \angle COA={90}^o$
$\angle AOE=\angle BOF$ (vertically opposite angles)
$\therefore \angle AOE={60}^o$
$\angle BOC+\angle BOD={180}^o$ (Linear pair)
$\angle BOC+{90}^o={180}^o$
$\therefore \angle BOC={90}^o$
$\angle BOC=\angle BOF+\angle COF$
${90}^o={60}^o+\angle COF$
$\therefore \angle COF={30}^o$
$\angle COF=\angle EOD$ (vertically opposite angles)
$\therefore \angle EOD={30}^o$