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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 848.
we have tan c=c/2c then tanc =0.5 c = shift tan(0.5)= 26.56
Draw a perpendicular BD from B to AC and the Let the two circles intersect in E.Now, AC = sqrt(5)*c while BD * AC = AB * BC or BD = (c * 2c)/(sqrt(5)*c) = 2c/sqrt(5)We can say that /_ACB//_ECB ~= BD/BE = 2/sqrt(5) or /_ACB = /_ECB * 2/sqrt(5) = 60/sqrt(5)or /_ ACB ~= 26.83281
I did not get this problemdear antonio gutierrez can u post this solnmy teachers the problem is immpossible
Dear Pranav,I apologize for the problem (Site Temporarily Unavailable), I am looking for a solution. In the meantime you can see this problem at http://agutie.homestead.com/files/geometry/p848-special-right-triangle-cathetus-1-2-angle-26.5-degrees.htm
i did not get the solutionplease help
Problem 848 Hints 1 Draw the cevian AD (D on AC) so that AD = DC2 Prove that triangle ABD is the special right triangle 3k-4k-5k (angles aprox. 37 and 53 degrees).3 Apply 180 degrees in isosceles triangle ADCGood luck