Look at the figure, O is the centre of in circle inscribed in\[\Delta \,PQR\]. |
A) \[95{}^\circ \]
B) \[125{}^\circ \]
C) \[115{}^\circ \]
D) \[125{}^\circ \]
Correct Answer: C
Solution :
(c): Make constructions as shown: By property of in circle, PO, QO and RO, and \[\angle \] bisectors \[\Rightarrow \] at point P, we have angles as \[25{}^\circ \] each; \[\angle OUP=90{}^\circ \] \[\therefore \] In \[\Delta \text{ }POU,\]\[\angle POU=90{}^\circ -25{}^\circ =65{}^\circ =y\] Now at point \[O,x+x+y+y+z+z=360{}^\circ \] \[\Rightarrow \]\[2\left( x+z \right)=360{}^\circ -2y-230{}^\circ \] \[\Rightarrow \]\[\angle QOR=x+z=115{}^\circ \]You need to login to perform this action.
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