Acute angle of a rhombus whose side is a mean proportional between it's diagonals is ?
Answers
Let the diagonal are x, y in length so
side^2 = xy
Now we know that diagonal bisect at 90 degrees so
side^2 = (x/2)^2 + (y/2)^2
xy = x^2 + y^2/4
x^2 + y^2 - 4xy = 0
x/y + y/x = 4
y/x + 1/y/x = 4
Let y/x = t
t^2 -4t + 1 = 0
t = 4+- underoot 12/2 = 2 +- underoot 3
or y/x = 2 +- underoot 3
Now let the angle is θ
tan θ = y/x = 2- underoot 3 [We have to find acute angle only]
Now tan 15 = tan(45-30) = 1/-1/underoot 3 / 1+ 1/underoot 3 = 2 - underoot 3
So tan θ = tan 15
Hence θ = 15
Refer to the attached image.
Consider a rhombus ABCD,
Let the length of diagonal AC = 'x' and length of diagonal BD = 'y'.
Let the side of a rhombus be 'a' meters.
It is given that side is a mean proportional between it's diagonals.
Therefore,
Since diagonals of rhombus are perpendicular bisectors.
Therefore,
Consider triangle BOC,
By pythagoras theorem,
Let
Consider triangle DOC,
Let
Consider
Since,
Rationalizing, we get
Therefore,
Consider,
On rationalizing, we get
Hence, the acute angle of rhombus whose side is a mean proportional between it's diagonals is 15 degrees.