Here an Isosceles triangle is inscribed in a square. The angle A is given by 15 degree...
prove that the triangle formed below is an equilateral triangle.
1
x(a/2) =tan15°
x=(1-√32)a
so, altitude of triangle=a-(1-√32)a=√3a2
so angle opposite to altitude =60°{by trignometry}
other angle =180-90-60=30
boththe 2 triangles are congruent
so all the angles are 60° each
so its an equilateral triangle
62wowo .. how did u get to know the value of tan 15 at sub class X level!
I dont even remember it now :P
btw there is a very nice geometric proof... (i Remember having done this one in class IX) :)
YOu will have to think of a very beasty construction.
1do we have to draw a circle around the equi triangle? then jon vertices with the circumcentre?
49hint: the vertex of the triangle drawn is equidistant from the base vertices....[1][1] i thnk ths is working,,,:)
36Nishant sir! please help me out in this question, it seems to be a wonderful one.
I proved the triangle to be isocelous and the other two triangles to be cong.
but could not prove that triangle to be equilateral. Some hints please....!!
36If we prove anyone of the two triangles to be an isocelous one then we r done.....
even dropping perpendicuclars and applying similarity dosen't work......
11nishant sir has said,u have to prove it in 9-10 std..so i used dis...
36@Soham -> okay....!!
well the only way out for this question is to draw a circle......
i googled this question and came up with the link having its solution.....
http://answers.yahoo.com/question/index?qid=20090419195053AAUPsYS
36
This envelope is something "Nishant sir" was talking about... :P
Such a small construction works out...!!
ABO ≡ MBO
so, LAOB = 75° or, AOD = 2 (105° - 75°) = 60°
Thus, AOD is equilateral....
lol after so long.... afterall i searched the solution
