Geometry Problem

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 848.

## Saturday, January 26, 2013

### Problem 848: Special Right triangle, Catheti or legs ratio 1:2, 26.5 Degrees. Double angle

Labels:
37-53,
angle,
special right triangle

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we have tan c=c/2c

ReplyDeletethen 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.

ReplyDeleteNow, 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 problem

ReplyDeletedear antonio gutierrez can u post this soln

my teachers the problem is immpossible

Dear Pranav,

DeleteI 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 solution

ReplyDeleteplease help

Problem 848 Hints

Delete1 Draw the cevian AD (D on AC) so that AD = DC

2 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 ADC

Good luck