A) \[\frac{2.05\sin 38{}^\circ }{\sin 42{}^\circ }\]
B) \[\frac{2.05\sin 42{}^\circ }{\sin 38{}^\circ }\]
C) \[\frac{2.05\cos 38{}^\circ }{\cos 42{}^\circ }\]
D) None of these
Correct Answer: A
Solution :
[a] \[\frac{\sin 38{}^\circ }{l}=\frac{\sin (SPO)}{2.05}\] \[=\frac{\sin (180{}^\circ -38{}^\circ -90{}^\circ -10{}^\circ )}{2.05}\] \[\Rightarrow l=\frac{2.05\sin 38{}^\circ }{\sin 42{}^\circ }\]You need to login to perform this action.
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