| In the given figure, \[AB\bot CD,\]\[AP||CD,\]\[\angle CBP=142{}^\circ .\]Find \[\angle ABP\]and \[\angle APB.\] |
 |
A) \[52{}^\circ ,\]\[38{}^\circ \]
B) \[56{}^\circ ,\]\[34{}^\circ \]
C) \[51{}^\circ ,\]\[39{}^\circ \]
D) \[57{}^\circ ,\]\[33{}^\circ \]
Correct Answer: A
Solution :
| \[\angle CBP+\angle PBD=180{}^\circ \] [linear pair ] |
| \[142{}^\circ +\angle PBD=180{}^\circ \] |
| \[\Rightarrow \] \[\angle PBD=180{}^\circ -142{}^\circ =38{}^\circ \] |
| Now, \[AP||CD\]and \[PB\] is the transversal. |
| \[\angle APB=\angle PBD\] [alternate angles] |
| \[\Rightarrow \] \[\angle APB=38{}^\circ \] \[[\because \angle PBD=38{}^\circ ]\] |
| \[\angle PAB+\angle ABD=180{}^\circ \] |
| [sum of the internal angles on the same side of the transversal] |
| \[\angle PAB+90{}^\circ =180{}^\circ \]\[[\because \angle ABD=90{}^\circ ]\] |
| \[\Rightarrow \]\[\angle PAB=180{}^\circ -90{}^\circ =90{}^\circ \] |
| In \[\Delta ABP,\]\[\angle ABP+\angle BPA+\angle PAB=180{}^\circ \] |
| \[\angle ABP+38{}^\circ +90{}^\circ =180{}^\circ \] |
| \[\Rightarrow \] \[\angle ABP=52{}^\circ \] |
| \[\therefore \]\[\angle ABP=52{}^\circ \] and \[\angle APB=38{}^\circ \] |
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