Saved Bookmarks
| 1. |
In the adjoining figure, AC is the diameter of the circle. If `angleBDC=115^(@)`, then find the value of `angle ACB`. |
|
Answer» square `ABDC` is a cyclic quadrilateral. `therefore angle BDC+BAC =180^@` `rArr115^@+angle BAC =180^@` `rArrangle ACB=180^@-115^@` ` 65^@` In `Delta ABC, AC` is the diameter of the circle. `thereforeangle ABC =90^@` Now, `angle ACB=180^@-(angle ABC +angle BAC) ` `=180^@=(90^@+65^@)` `rArrangle ACB =25^@` . |
|