A) \[\text{20 }\!\!{}^\circ\!\!\text{ }\]
B) \[\text{35 }\!\!{}^\circ\!\!\text{ }\]
C) \[\text{45 }\!\!{}^\circ\!\!\text{ }\]
D) \[\text{60 }\!\!{}^\circ\!\!\text{ }\]
Correct Answer: C
Solution :
Since OS is normal to the mirror AB. \[\therefore \angle {{i}_{1}}=45{}^\circ ,\] Also \[\angle {{r}_{1}}=45{}^\circ \] In \[\Delta CQR,\angle C=90{}^\circ ,\angle CQR=45{}^\circ \therefore \angle CRQ=45{}^\circ \] [\[\because \]Sum of all the three angles in triangle is\[\text{180 }\!\!{}^\circ\!\!\text{ }\].] Hence, \[\angle {{\text{i}}_{2}}=45{}^\circ \] [TR is the normal to the mirror BD] \[\therefore \angle {{r}_{2}}=45{}^\circ \]You need to login to perform this action.
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