Question about angles of rhoumbus

Question

Problem:

Consider a rhombus (Diamond) such that each of its side is the geometric mean of its diameters. I mean if length of each side is X and the diameters a and b; then $X^2$ = a.b

Find the angles of rhombus.

Answer 1

The area of shape is $\dfrac{ab}{2}=\dfrac{x^2}{2}$.

Also the area of $\triangle ADC$ and $\triangle ABC$ is equal to $\dfrac{x^2}2\cdot\sin D=\dfrac{x^2}2\cdot\sin B$, so $$S_{ABCD}=\frac{x^2}{2}=(\dfrac{x^2}2\cdot\sin D+\dfrac{x^2}2\cdot\sin B)=x^2\cdot\sin D=x^2\cdot\sin B\\[4ex]\Rightarrow\sin D =\sin B=\frac12\\[4ex]\Rightarrow D=B=30^{\circ}$$ And in the conclusion $$A=C=150^\circ$$

Answer 2

Let's say that we have the rhombus as in the image and assume that $a>b$.

Now, since $\triangle AMB$ is right-angled, we have that: $$AB^2 = AM^2 + BM^2 \iff ab = \left(\frac{a}{2}\right)^2 + \left(\frac{b}{2}\right)^2 \iff ab = \frac{a^2}{4} + \frac{b^2}{4} \iff a^2 -4ab +b^2 =0 .$$ From the equation above it holds that $a = (2+\sqrt{3})b$ (there is also another solution $a = (2-\sqrt{3})b < b$, thus rejected, since we assumed that $a>b$).

Setting $\hat {BAM} = \theta$, we have that $$\tan\theta = \dfrac{b/2}{a/2} =\frac{b}{a} = \frac{1}{2+\sqrt{3}}.$$

Now, we have that $\hat{BAD} = 2\theta$, thus: $$\tan(2\theta) = \frac{2\tan\theta}{1-\tan^2\theta} =\cdots = \frac{\sqrt{3}}{3} \implies 2\theta =\hat{BAD}= 30^\circ.$$