In the question we are given a rhombus. We are also told that one of the diagonals of the rhombus is equal to the geometric mean of one of its side length and length the length of other diagonal.
I am trying to find the measure of all the angles of the rhombus. (opposite angles are equal)
What I tried.
Let the length of 2 diagonals be a and b. Let the side length of rhombus be c. Since diagonals bisect each other at right angles then
a^2 + b^c = 4c^2 -------- eq (1)
According to given condition
a = (bc)^(1/2) ---------- eq (2)
After this i am stuck. I tried using different area formulas of rhombus and equated them a lot and got answer as 0.
I am not able to solve further yet i am trying. Thanks for the help and your time.
Guide me.
Best Answer
As per your working, $a^2+b^2 = 4c^2, a = \sqrt{bc}$
$ \implies b^2 + bc = 4c^2$
$ \displaystyle (b + \frac{c}{2})^2 = \frac{17c^2}{4}$
$ \displaystyle \frac{b}{c} = \frac{-1 \pm \sqrt{17}}{2}$
Now if $A$ and $B$ are two adjacent vertices of rhombus and $O$ is the intersection of diagonals, $\triangle OAB$ is a right angled triangle. Cosine of one of the angles of the triangle will be $\displaystyle \frac{b/2}{c}$.
As a diagonal bisects the vertex angles in the rhombus, one of the angles of the rhombus will be,
$\theta = 2 \arccos (\frac{b}{2c})$ and so the other angle will be $(\pi-\theta)$.
As you know $\frac{b}{c}$, you can now find the angle. Note that you will discard $ \displaystyle \frac{b}{c} = \frac{-1 - \sqrt{17}}{2}$ as angle between diagonal and side is less than $90^0$.