A) 12\[\sqrt{2}\] cm
B) 16 cm
C) 12 cm
D) 8 cm
Correct Answer: C
Solution :
Here, \[\angle PQN={{45}^{o}}\] So, \[PN=h\,\tan \,{{45}^{o}}=32m\] Now, \[\mu =\frac{\sin i}{\sin r}\] So, \[\sin r=\frac{sin\,i}{\mu }\] or \[\sin r=\frac{\sin \,{{45}^{o}}}{(4/3)}=\frac{3}{4\sqrt{2}}\] \[\tan r=\frac{\sin r}{\cos r}\] \[=\frac{\sin \,r}{\sqrt{1-{{\sin }^{2}}r}}\] \[=\frac{3/4\sqrt{2}}{\sqrt{1-\frac{9}{32}}}=\frac{3}{\sqrt{23}}\] So, ON= h tan r \[=32\times \frac{3}{\sqrt{23}}=20cm\] \[OP=PN-ON\] \[=32-20=12\text{ }cm\]You need to login to perform this action.
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