A) \[\theta ={{\tan }^{-1}}(\tan {{\delta }_{1}}\tan {{\delta }_{2}})\]
B) \[\theta ={{\tan }^{-1}}(\tan {{\delta }_{1}}+\tan {{\delta }_{2}})\]
C) \[\theta ={{\tan }^{-1}}\left( \frac{\tan {{\delta }_{1}}}{\tan {{\delta }_{2}}} \right)\]
D) \[\theta ={{\tan }^{-1}}(\tan {{\delta }_{1}}-\tan {{\delta }_{2}})\]
Correct Answer: C
Solution :
Here,\[\tan {{\delta }_{1}}=\frac{v}{H\cos \theta }\] \[\tan {{\delta }_{2}}=\frac{v}{H\cos ({{90}^{o}}-\theta )}=\frac{v}{H\sin \theta }\] \[\frac{\tan {{\delta }_{1}}}{\tan {{\delta }_{2}}}=\frac{\sin \theta }{\cos \theta }=\tan \theta \] or \[\theta ={{\tan }^{-1}}\left( \frac{\tan {{\delta }_{1}}}{\tan {{\delta }_{2}}} \right)\]You need to login to perform this action.
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