A) \[\sqrt{2}+1\]
B) \[2-\sqrt{3}\]
C) \[\sqrt{2}-1\]
D) \[2+\sqrt{3}\]
Correct Answer: B
Solution :
[b] In \[\Delta ABC,\] \[x=h\,\,\cot 2\theta \] ...(i) In \[\Delta ABD,\] \[h=(2h+x)\,\tan \theta \] ...(ii) Putting value of x from (i) into (ii), we get \[1=\left( 2+\frac{1}{\tan 2\theta } \right)\tan \theta \] \[\Rightarrow \,\,\,\,\,\,\,\,\,1=\left( 2+\frac{1-{{\tan }^{2}}\theta }{2\tan \theta } \right)\tan \theta \] \[\therefore \,\,\,\,\,{{\tan }^{2}}\theta -4\tan \theta +1=0\] \[\therefore \,\,\,\,\,\tan \theta =2-\sqrt{3}\] (as \[\tan \theta =2+\sqrt{3}\] not possible)You need to login to perform this action.
You will be redirected in
3 sec