The sides of a triangle are sin α, cos α and √(1 + sin α cos α) for some 0

Q: Why do we square the sides to compare them?

Comparing square roots directly can be messy and prone to error. Squaring is a monotonic transformation for positive numbers, meaning if $x > y > 0$, then $x^2 > y^2$. By squaring, we remove the radical and make the comparison straightforward.

Q: Can I use the Sine Rule instead?

You could, but the Sine Rule requires knowing the angles or using the circumradius $R$. The Cosine Rule is more direct when you have all three sides, as it relates the sides directly to the cosine of the angle.

Q: Why is the angle $120^\circ$ and not $60^\circ$?

The cosine value we calculated is $-1/2$. In the range $(0, 180^\circ)$, the cosine function is negative only in the second quadrant. The angle whose cosine is $-1/2$ is $120^\circ$. If it were $60^\circ$, the cosine would be positive $1/2$.

Question

The sides of a triangle are $\sin \alpha, \cos \alpha$ and $\sqrt{1 + \sin \alpha \cos \alpha}$ for some $0 < \alpha < \frac{\pi}{2}$. Then the greatest angle of the triangle is

Options:

(A) $150^\circ$

(B) $90^\circ$

(C) $120^\circ$

(D) $60^\circ$

Sigma Percentile · Accepted Answer

### Analyzing the Setup We are given a triangle with side lengths defined by the angle $\alpha$, where $0 < \alpha < \frac{\pi}{2}$. The sides are: $a = \sin \alpha$ $b = \cos \alpha$ $c = \sqrt{1 + \sin \alpha \cos \alpha}$ Since $\alpha$ lies in the first quadrant, both $\sin \alpha$ and $\cos \alpha$ are positive, ensuring the triangle is well-defined. ### The Quest for the Longest Side In any triangle, the greatest angle is always located opposite the longest side. To identify the longest side, we compare the squares of the side lengths: $a^2 = \sin^2 \alpha$ $b^2 = \cos^2 \alpha$ $c^2 = 1 + \sin \alpha \cos \alpha$ We know that $\sin^2 \alpha < 1$ and $\cos^2 \alpha < 1$. Because $\sin \alpha \cos \alpha > 0$ for $0 < \alpha < \frac{\pi}{2}$, it follows that $c^2 = 1 + \sin \alpha \cos \alpha > 1$. Thus, $c^2$ is strictly greater than both $a^2$ and $b^2$. This confirms that $c$ is the longest side, and the angle $C$ opposite to it is the greatest angle of the triangle. ### The Power of the Cosine Rule To determine the value of angle $C$, we employ the Law of Cosines: $$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$ Substituting our expressions for $a$, $b$, and $c$ into this formula, we obtain: $$\cos C = \frac{\sin^2 \alpha + \cos^2 \alpha - (1 + \sin \alpha \cos \alpha)}{2 \sin \alpha \cos \alpha}$$ ### The Elegance of Algebraic Cancellation We utilize the fundamental trigonometric identity $\sin^2 \alpha + \cos^2 \alpha = 1$ to simplify the numerator. The expression becomes: $$\cos C = \frac{1 - (1 + \sin \alpha \cos \alpha)}{2 \sin \alpha \cos \alpha}$$ Distributing the negative sign yields: $$\cos C = \frac{1 - 1 - \sin \alpha \cos \alpha}{2 \sin \alpha \cos \alpha}$$ The constants $1$ and $-1$ cancel out, leaving: $$\cos C = \frac{-\sin \alpha \cos \alpha}{2 \sin \alpha \cos \alpha}$$ Since $\sin \alpha eq 0$ and $\cos \alpha eq 0$ in the first quadrant, we cancel the common term $\sin \alpha \cos \alpha$ to find: $$\cos C = -\frac{1}{2}$$ ### The Final Revelation We seek the angle $C$ such that $\cos C = -\frac{1}{2}$, with the constraint that $0^\circ < C < 180^\circ$. The cosine function is negative in the second quadrant, specifically at $120^\circ$. Therefore, the greatest angle of the triangle is **$120^\circ$** (or $\frac{2\pi}{3}$ radians).

Animated Solution for Mathematics - Trigonometry: The sides of a triangle are $sin α, cos α$ and $1 + sin α cos α$ for some $0 < α < \frac{π}{2}$ . Then the greatest angle of the triangle is

Select Answer:

Visualized Solution

Define the Sides of the Triangle

Analyze the Range of $sin α$ and $cos α$

Compare the Squares of the Sides

Establish that $c$ is the Longest Side

Identify the Greatest Angle $C$

Apply the Cosine Rule

Substitute the Side Lengths

Simplify using $sin^{2} α + cos^{2} α = 1$

Expand the Numerator

Cancel the Common Terms

Calculate the Greatest Angle $C$

Final Answer and Summary

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Quest for the Longest Side

The Power of the Cosine Rule

The Elegance of Algebraic Cancellation

The Final Revelation

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sides of a triangle are sinα,cosα and 1+sinαcosα​ for some 0<α<2π​. Then the greatest angle of the triangle is