A) \[{{\tan }^{-1}}\left( \frac{3}{5} \right)\]
B) \[{{\tan }^{-1}}\left( \frac{4}{5} \right)\]
C) \[\pi \]
D) \[\frac{\pi }{2}\]
Correct Answer: A
Solution :
Tangent at (16,8) to both are \[8y=2(x+16)\] ..... (i) and \[16x=16(y+8)\] .....(ii) \ \[{{m}_{1}}=\frac{1}{4},\,\,{{m}_{2}}=1\] \[\tan \theta =\frac{{{m}_{2}}-{{m}_{1}}}{1+{{m}_{2}}{{m}_{1}}}=\left( \frac{3}{5} \right)\]Þ \[\theta ={{\tan }^{-1}}\left( \frac{3}{5} \right)\]. Aliter: Using direct formula \[\theta ={{\tan }^{-1}}\frac{3{{a}^{1/3}}{{b}^{1/3}}}{2({{a}^{2/3}}+{{b}^{2/3}})},\] where \[a=1\] and \[b=8\] \[={{\tan }^{-1}}\frac{6}{2(1+4)}={{\tan }^{-1}}\frac{3}{5}\].
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