MHB Right Angled Triangle: Find $\sin(\theta)$

[kaliprasad]

Let $a,b,c$ be the sides of a right angled triangle. Let $\theta$ be the smallest angle of this triangle.

If $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ are also the sides of a right angled triangle then show that $\sin(\theta) = \dfrac{\sqrt{5} - 1}{2}$

 

[Sudharaka]

Let $a,b,c$ be the sides of a right angled triangle. Let $\theta$ be the smallest angle of this triangle.

If $\dfrac{1}{a}, \dfrac{1}{b}, \dfrac{1}{c}$ are also the sides of a right angled triangle then show that $\sin(\theta) = \dfrac{\sqrt{5} - 1}{2}$

Interesting question. :)

Spoiler

Without loss of generality let us take $c$ to be the length of the hypotenuse and $b$ to be the length of the smallest side. That is, \(c>a>b\). Therefore \(\frac{1}{c}<\frac{1}{a}<\frac{1}{b}\). That is the \(\frac{1}{b}\) is the length of the hypotenuse of the smaller triangle. Using the Pythagorean theorem for both triangles we get,

\[a^2+b^2=c^2\]

and

\[\left(\frac{1}{a}\right)^2+\left(\frac{1}{c}\right)^2=\left(\frac{1}{b}\right)^2\]

From these two equations we can get,

\[bc=a^2\]

And hence,

\[\sin\theta=\frac{b}{c}=\frac{a^2}{c^2}=cos^2\theta=1-\sin^2\theta\]

\[\therefore \sin^2\theta+\sin\theta-1=0\Rightarrow \sin\theta=\frac{\sqrt{5}-1}{2}\]