Saved Bookmarks
| 1. |
In the adjoining , `O` is the centre of the circle. `ACB` is a segment. If `angle OAB=30^(@)`, then find the value of `angle ACB`. |
|
Answer» In `Delta OAB`, `AO=BO` (radii of circle) `rArrangle ABO=angleBAO` `rArrangle ABO=30^@` Now, `angle AOB=180^@-(angleABO+angle BAO)` `=180^@-(30^@+30^@)` `120^@` Arc AB subtends angles at centre `=angle AOB` remaining circle `=angle ACB` `therefore angle ACB=(1)/(2)angle AOB` `=(1)/(2)xx120^@` `=60^@` |
|