A) 0.9659
B) 2
C) 1
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
\[\frac{P}{\sin {{\theta }_{1}}}=\frac{Q}{\sin {{\theta }_{2}}}=\frac{R}{\sin \,\,150{}^\circ }\] \[\Rightarrow \,\frac{1.93}{\sin {{\theta }_{1}}}=\frac{R}{\sin \,150{}^\circ }\] \[\Rightarrow \,R=\frac{1.93\times \sin \,150{}^\circ }{\sin {{\theta }_{1}}}=\frac{1.93\times 0.5}{0.9659}=1\]You need to login to perform this action.
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