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

  • Thread starter Thread starter kaliprasad
  • Start date Start date
  • Tags Tags
    Triangle
AI Thread Summary
In a right-angled triangle with sides \(a, b, c\) and the smallest angle \(\theta\), it is established that if the reciprocals \(\frac{1}{a}, \frac{1}{b}, \frac{1}{c}\) also form a right-angled triangle, then \(\sin(\theta)\) can be expressed as \(\frac{\sqrt{5} - 1}{2}\). This relationship highlights a unique property of the triangle's angles and side lengths. The derivation involves understanding the geometric implications of both triangles. The discussion emphasizes the mathematical connection between the angles and the ratios of the sides in right-angled triangles.
kaliprasad
Gold Member
MHB
Messages
1,333
Reaction score
0
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 $\sin(\theta) = \dfrac{\sqrt{5} - 1}{2}$
 
Mathematics news on Phys.org
kaliprasad said:
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. :)

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,

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

\[\therefore \sin^2\theta+\sin\theta-1=0\Rightarrow \sin\theta=\frac{\sqrt{5}-1}{2}\]
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Back
Top