| In the given figure, \[AM\bot BC\] and AN is the bisector of \[\angle A.\] If \[\angle ABC=70{}^\circ \]and \[\angle ACB=20{}^\circ ,\] then \[\angle MAN\] is equal to |
 |
A) \[20{}^\circ \]
B) \[25{}^\circ \]
C) \[15{}^\circ \]
D) \[30{}^\circ \]
Correct Answer: B
Solution :
| In \[\Delta ABC,\]\[\angle B+\angle C+\angle A=180{}^\circ \] |
| \[\Rightarrow \] \[\angle A=180{}^\circ -90{}^\circ =90{}^\circ \] |
| But is bisector of \[\angle A.\] |
| \[\therefore \] \[\angle NAC=\angle NAB=45{}^\circ \] |
| In \[\Delta ANC,\] |
| \[\angle ANC=180{}^\circ -(20{}^\circ +\angle NAC)\] |
| \[=180{}^\circ -(20{}^\circ +45{}^\circ )\] |
| \[=180{}^\circ -65{}^\circ =115{}^\circ \] |
| \[\therefore \] \[\angle ANM=180{}^\circ -115{}^\circ =65{}^\circ \] |
| \[\angle MAN=180{}^\circ -(90{}^\circ +65{}^\circ )\] |
| \[=180{}^\circ -155{}^\circ =25{}^\circ \] |
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