A) 3
B) \[\frac{4}{3}\]
C) \[\frac{1}{2}\]
D) 2
Correct Answer: D
Solution :
\[A(-16,0)\] \[B(16+2(4),0)=B(24,0)\] \[AC=CB\Rightarrow C(4,0)\] C is the cimcumcentre \[BP=\sqrt{{{8}^{2}}+{{16}^{2}}}=8\sqrt{5}\] \[PC=\sqrt{{{12}^{2}}+{{16}^{2}}}=20\] \[PC=\sqrt{{{12}^{2}}+{{16}^{2}}}=20\] \[\Rightarrow \tan \theta =2\]You need to login to perform this action.
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