A) \[\sqrt{3},\sqrt{2}\]
B) \[\sqrt{3},-\sqrt{3}\]
C) \[-\sqrt{5},\sqrt{3}\]
D) \[\sqrt{2},-\sqrt{2}\]
E) \[\sqrt{5},-\sqrt{5}\]
Correct Answer: B
Solution :
Since, triangle is equilater. \[\therefore \] \[\angle ACB=60{}^\circ \] \[\therefore \] \[\angle ACD=30{}^\circ \] Now, in right angle triangle\[\Delta ADC,\] \[\tan 30{}^\circ =\frac{1}{x}\] \[\Rightarrow \] \[\frac{1}{\sqrt{3}}=-\frac{1}{x}\] \[\Rightarrow \] \[x=-\sqrt{3}\] Similarly, in right angle triangle BDC \[\tan 30=-\frac{1}{x}\] \[\Rightarrow \] \[\frac{1}{\sqrt{3}}=-\frac{1}{x}\] \[\Rightarrow \] \[x=-\sqrt{3}\]You need to login to perform this action.
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