| In the figure, \[\angle CAB=72{}^\circ ,\]\[\angle CBA=74{}^\circ \]and \[\angle CED=112{}^\circ \]Find \[\angle CDE.\] |
 |
A) \[34{}^\circ \]
B) \[33{}^\circ \]
C) \[35{}^\circ \]
D) \[38{}^\circ \]
Correct Answer: A
Solution :
| In \[\Delta ABC,\]\[\angle CAB+\angle ABC+\angle BCA=180{}^\circ \] |
| \[72{}^\circ +74{}^\circ +\angle BCA=180{}^\circ \] |
| \[146{}^\circ +\angle BCA=180{}^\circ \] |
| \[\angle BCA=34{}^\circ \] |
| Now, \[\angle BCA=\angle DCE\][vertically opposite angle] |
| In \[\Delta CDE,\] |
| \[\angle CED+\angle CDE+\angle DCE=180{}^\circ \] |
| \[112{}^\circ +\angle CDE+34{}^\circ =180{}^\circ \] |
| \[\Rightarrow \] \[\angle CDE=180{}^\circ -\,\,(112+34){}^\circ \] |
| \[\Rightarrow \] \[\angle CDE=180{}^\circ -146{}^\circ =34{}^\circ \] |
| \[\Rightarrow \] \[\angle CDE=34{}^\circ \] |
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