| In \[\Delta ABC,\]\[\angle A:\angle B:\angle C=2:3:4,\] a line CD drawn perpendicular to AB, then \[\angle ACD\] is [SSC (CGL) 2013] |
A) \[80{}^\circ \]
B) \[20{}^\circ \]
C) \[40{}^\circ \]
D) \[60{}^\circ \]
Correct Answer: C
Solution :
| Let the angles be \[2x,\]\[3x\]and\[4x.\] |
| Since, the sum of interior angles of triangle is \[180{}^\circ .\] |
| Then, \[2x+3x+4x=180{}^\circ \] |
| \[\Rightarrow \] \[9x=180{}^\circ \] |
| \[\therefore \] \[x=20{}^\circ \] |
| Now, \[\angle A=2x=2\times 20{}^\circ =40{}^\circ \] |
| \[\angle B=3x=3\times 20{}^\circ =60{}^\circ \] |
| \[\angle C=4x=4\times 20{}^\circ =80{}^\circ \] |
| Now, \[AB\bot CD\]and AC be the transversal. |
| Then, \[\angle BCD=\angle ACD\][alternate interior angles] |
| \[\therefore \] \[\angle ACD=40{}^\circ \] |
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