Let in a right angled triangle, the smallest angle be θ. If a triangle formed by taking the reciproc

Q: Wait, why is $\frac{1}{a}$ the hypotenuse? Shouldn't it be $\frac{1}{c}$ because $c$ was the original hypotenuse?

That is a classic trap! Remember, the hypotenuse is defined by the *longest* side, not by its original label. Since $a$ is the smallest side in the original triangle, its reciprocal $\frac{1}{a}$ is the largest value. Therefore, in the new triangle, $\frac{1}{a}$ must be the hypotenuse.

Q: Why did we reject the root $\frac{3 + \sqrt{5}}{2}$ when solving the quadratic equation?

Great observation. We set $x = \sin^2 \theta$. Since $\sin \theta$ is between $0$ and $1$, $\sin^2 \theta$ must also be between $0$ and $1$. The value $\frac{3 + \sqrt{5}}{2}$ is approximately $\frac{3 + 2.236}{2} \approx 2.618$, which is greater than $1$. Thus, it is physically impossible for $\sin^2 \theta$ to take that value.

Q: How did you know $6 - 2\sqrt{5}$ was $(\sqrt{5} - 1)^2$?

It comes from the algebraic identity $(x - y)^2 = x^2 + y^2 - 2xy$. If we let $x = \sqrt{5}$ and $y = 1$, then $(\sqrt{5} - 1)^2 = (\sqrt{5})^2 + 1^2 - 2(\sqrt{5})(1) = 5 + 1 - 2\sqrt{5} = 6 - 2\sqrt{5}$. Recognizing these perfect squares is a vital skill for simplifying radical expressions in JEE problems.

Question

Let in a right angled triangle, the smallest angle be $	heta$. If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then $\sin 	heta$ is equal to:

Options:

(A) $\frac{\sqrt{5}+1}{4}$

(B) $\frac{\sqrt{5}-1}{2}$

(C) $\frac{\sqrt{2}-1}{2}$

(D) $\frac{\sqrt{5}-1}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a right-angled triangle with sides $a$, $b$, and $c$, where $c$ is the hypotenuse. We are told the smallest angle is $	heta$.

This implies that the side opposite to this angle, $a$, must be the shortest side. Consequently, we have the inequality $a < b < c$.

The problem introduces a new triangle formed by the reciprocals of these sides: $\frac{1}{a}$, $\frac{1}{b}$, and $\frac{1}{c}$. We are given that this new triangle is also right-angled.

### The Transformation

When we take the reciprocals, the hierarchy of the sides flips. Since $a$ was the smallest, $\frac{1}{a}$ becomes the largest.

In any right-angled triangle, the longest side is the hypotenuse. Therefore, in our new triangle, $\frac{1}{a}$ must be the hypotenuse.

Applying the Pythagorean theorem to this new set of sides, we obtain:

$$ \left( \frac{1}{a} ight)^2 = \left( \frac{1}{b} ight)^2 + \left( \frac{1}{c} ight)^2 $$

This simplifies to the following relationship:

$$ \frac{1}{a^2} = \frac{1}{b^2} + \frac{1}{c^2} $$

### Bridging Geometry and Trigonometry

Now, we relate this to the original triangle. We know that $a = c \sin 	heta$ and $b = c \cos 	heta$.

Substituting these into our equation, we get:

$$ \frac{1}{c^2 \sin^2 	heta} = \frac{1}{c^2 \cos^2 	heta} + \frac{1}{c^2} $$

Since the hypotenuse $c$ appears in every term, we can divide the entire equation by $\frac{1}{c^2}$ to eliminate it. This leaves us with a purely trigonometric identity:

$$ \frac{1}{\sin^2 	heta} = \frac{1}{\cos^2 	heta} + 1 $$

### The Algebraic Battle

To solve for $\sin 	heta$, we first combine the right side of the equation:

$$ \frac{1}{\sin^2 	heta} = \frac{1 + \cos^2 	heta}{\cos^2 	heta} $$

Cross-multiplying yields $\cos^2 	heta = \sin^2 	heta (1 + \cos^2 	heta)$. To solve this, we express everything in terms of $\sin 	heta$ using the identity $\cos^2 	heta = 1 - \sin^2 	heta$:

$$ 1 - \sin^2 	heta = \sin^2 	heta (1 + 1 - \sin^2 	heta) $$

Expanding this expression results in $1 - \sin^2 	heta = 2\sin^2 	heta - \sin^4 	heta$. Rearranging all terms to one side, we arrive at the quadratic form:

$$ \sin^4 	heta - 3\sin^2 	heta + 1 = 0 $$

### The Final Resolution

Let $x = \sin^2 	heta$. We now have the quadratic equation $x^2 - 3x + 1 = 0$.

Using the quadratic formula, we find:

$$ x = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} $$

We must reject the root $\frac{3 + \sqrt{5}}{2}$ because it exceeds $1$, which is impossible for $\sin^2 	heta$. Thus, we have $\sin^2 	heta = \frac{3 - \sqrt{5}}{2}$.

Taking the square root, we get:

$$ \sin 	heta = \sqrt{\frac{3 - \sqrt{5}}{2}} = \sqrt{\frac{6 - 2\sqrt{5}}{4}} $$

Recognizing that $6 - 2\sqrt{5} = (\sqrt{5} - 1)^2$, we finally arrive at the result:

$$ \sin 	heta = \frac{\sqrt{5} - 1}{2} $$

This value is the reciprocal of the **Golden Ratio**. You have successfully navigated the geometry, trigonometry, and algebra to reach the final answer.

Animated Solution for Mathematics - Trigonometry: Let in a right angled triangle, the smallest angle be $θ$ . If a triangle formed by taking the reciprocal of its sides is also a right angled triangle, then $sin θ$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Triangle

Defining Sides via Trigonometry

Ordering the Reciprocal Sides

Applying Pythagoras Theorem

Substitution of Trig Ratios

Simplifying the Equation

Converting to Sine Terms

Forming the Quadratic Equation

Solving the Quadratic

Finding Sine Theta

Final Conclusion

The Sigma Insight: Trigonometric Ratios and Identities

Analyzing the Setup

The Transformation

Bridging Geometry and Trigonometry

The Algebraic Battle

The Final Resolution

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