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adobe offline activation response code adobe offline activation response codeand many more benefits! Find us on Facebook GMAT Club Timer Informer Hi GMATClubber! Thank you for using the timer! We noticed you are actually not timing your practice. Click the START button first next time you use the timer. There are many benefits to timing your practice, including: Show Tags 03 Oct 2012, 11:12 [revised] Show Tags 03 Oct 2012, 12:44 ryancord wrote: DibyaK wrote: I have a doubt please check this for me… In a triangle ABC with AC as side of square and DF as the angle at A.AHG is the height of the square i.e. FGH. I also know DG is perpendicular to HG, HF is perpendicular to GH and AB is perpendicular to DF.Also AB is the side of rectangle ABCD.In this rectagle, AB represents ADC, CD represents BDA and AD represents CDA. The length of the diagonal of rectangle ABCD is 48.Find the area of the traingle: Please note the angles at A as compared to the one indicated on the slide. Please do the calculations also, show me the slide and tell me how to calculate it. _________________ Show Tags (1) $\frac{B}{AD} = \frac{2}{1}=2$. Hence, $B = 2AD$.(2) From the earlier post, show that $\tan \theta_1 = \tan \theta_2 = \tan 2 = \sqrt 2$.(3) $AD = 2\tan \theta = 4\tan 2 = 4\sqrt 2$. Hence, $AD^2 = 32\tan 2 = 32\sqrt 2 = 48$.(4) Since $\theta = 2\theta_1 = 2\pi – 2\theta_2$, it follows that $2\theta = 2\pi – 2(2\pi – 2\theta_2) = 2\pi – 2(2\pi – 2\theta_2) = 2\pi – 4\theta_2$. Hence, \$\tan 2\theta = 2\tan 2\theta_2 = 2\sq f30f4ceada