Let PQRS be a quadrilateral in a plane, where QR = 1, ∠PQR = ∠QRS = 70°, ∠PQS = 15° and ∠PRS = 40°.

Q: Wait, why did we assume $\triangle QRS$ is isosceles?

We didn't assume it! We calculated the angles: $\angle RQS = 70^\circ - 15^\circ = 55^\circ$ and $\angle QSR = 180^\circ - (70^\circ + 55^\circ) = 55^\circ$. Since two angles are equal, the triangle must be isosceles by the converse of the Isosceles Triangle Theorem.

Q: Why can't we just use the Law of Cosines?

You absolutely could, but it would lead to a very messy algebraic path. The Sine Rule is much more elegant here because it allows us to work directly with the ratios of sides and the sines of the opposite angles, which simplifies beautifully with the given trigonometric values.

Q: How do I know which interval to pick for $\sec 40^\circ$?

Great question! We know that $30^\circ < 40^\circ < 45^\circ$. Since the secant function is strictly increasing in the first quadrant, we have $\sec 30^\circ < \sec 40^\circ < \sec 45^\circ$. This gives us $\frac{2}{\sqrt{3}} < \sec 40^\circ < \sqrt{2}$, which is approximately $1.15 < \sec 40^\circ < 1.414$. This value clearly falls within the intervals $(0, \sqrt{2})$ and $(1, 2)$.

Question

Let $PQRS$ be a quadrilateral in a plane, where $QR = 1$, $\angle PQR = \angle QRS = 70^\circ$, $\angle PQS = 15^\circ$ and $\angle PRS = 40^\circ$. If $\angle RPS = 	heta^\circ$, $PQ = \alpha$ and $PS = \beta$, then the interval(s) that contain(s) the value of $4\alpha\beta \sin 	heta^\circ$ is/are

Options:

(A) $(0, \sqrt{2})$

(B) $(1, 2)$

(C) $(\sqrt{2}, 3)$

(D) $(2\sqrt{2}, 3\sqrt{2})$

Sigma Percentile · Accepted Answer

### The Geometry of Hidden Symmetry

Imagine you are standing on a flat plane, looking at a quadrilateral $PQRS$. At first glance, it seems like just another collection of lines and angles.

As an engineer or a physicist, you know that geometry is rarely just about shapes; it is about the hidden relationships that govern the universe. Let us peel back the layers of this problem together.

### Phase 1: Unlocking $	riangle QRS$

We start with the base $QR = 1$. We are given $\angle PQR = 70^\circ$ and $\angle QRS = 70^\circ$.

Our first mission is to understand the triangle $QRS$. We know $\angle PQS = 15^\circ$, so the remaining part of the angle at $Q$ within $	riangle QRS$ is:
$$\angle RQS = \angle PQR - \angle PQS = 70^\circ - 15^\circ = 55^\circ$$

Now, look at the sum of angles in $	riangle QRS$. We have $\angle QRS = 70^\circ$ and $\angle RQS = 55^\circ$. The third angle, $\angle QSR$, must be:
$$\angle QSR = 180^\circ - (70^\circ + 55^\circ) = 55^\circ$$

Suddenly, the beauty of the problem reveals itself: $	riangle QRS$ is an isosceles triangle because $\angle RQS = \angle QSR = 55^\circ$. This means the side $RS$ must be equal to the base $QR$, so $RS = 1$. We have unlocked our first key.

### Phase 2: The Sine Rule Dance

Now, let us shift our focus to $	riangle PQR$. We know $\angle QRS = 70^\circ$ and $\angle PRS = 40^\circ$, so:
$$\angle PRQ = 70^\circ - 40^\circ = 30^\circ$$

In $	riangle PQR$, the angle sum property gives us $\angle QPR = 180^\circ - (70^\circ + 30^\circ) = 80^\circ$. Applying the Sine Rule in $	riangle PQR$, we get:
$$\frac{PQ}{\sin 30^\circ} = \frac{QR}{\sin 80^\circ}$$

Since $QR = 1$ and $\sin 30^\circ = \frac{1}{2}$, we find $\alpha = PQ$ as:
$$\alpha = PQ = \frac{1}{2 \sin 80^\circ}$$

Next, we turn to $	riangle PRS$. Applying the Sine Rule here, we have:
$$\frac{PS}{\sin 40^\circ} = \frac{RS}{\sin 	heta^\circ}$$

Since $RS = 1$ and $PS = \beta$, we get $\beta \sin 	heta^\circ = \sin 40^\circ$. This is a powerful relation!

### Phase 3: The Grand Simplification

We are asked to evaluate $4\alpha\beta \sin 	heta^\circ$. Substituting our findings, we have:
$$4 \left( \frac{1}{2 \sin 80^\circ} ight) (\sin 40^\circ) = \frac{2 \sin 40^\circ}{\sin 80^\circ}$$

This is where the magic happens. Using the double-angle identity $\sin 80^\circ = 2 \sin 40^\circ \cos 40^\circ$, the expression becomes:
$$\frac{2 \sin 40^\circ}{2 \sin 40^\circ \cos 40^\circ} = \frac{1}{\cos 40^\circ} = \sec 40^\circ$$

The complexity vanishes, leaving us with a simple secant function.

### Conclusion: The Final Interval

Finally, we evaluate $\sec 40^\circ$. Since $30^\circ < 40^\circ < 45^\circ$, we know that $\sec 30^\circ < \sec 40^\circ < \sec 45^\circ$.

This implies:
$$\frac{2}{\sqrt{3}} < \sec 40^\circ < \sqrt{2}$$

Numerically, this is approximately $1.15 < \sec 40^\circ < 1.414$. Looking at our options, this value fits perfectly within the intervals $(0, \sqrt{2})$ and $(1, 2)$.

You have successfully navigated the geometry and trigonometry of this quadrilateral. Keep this analytical mindset, and no problem will ever be too daunting!

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Visualized Solution

$Given Quadrilateral P QR S$

$Calculate ∠ R QS$

$Calculate ∠ QS R$

$Sides of Isosceles △ QR S$

$Calculate ∠ P R Q$

$Calculate ∠ QP R$

$Sine Rule in △ P QR$

$Find α$

$Sine Rule in △ P R S$

$Relation for β$

$Evaluate 4 α β sin θ^{\circ}$

$Simplify Expression$

$Find the Interval$

The Sigma Insight: Properties of Triangles

The Geometry of Hidden Symmetry

Phase 1

Phase 2

Phase 3

Conclusion

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Quadrilateral PQRS

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculate ∠RQS

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculate ∠QSR

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sides of Isosceles △QRS

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculate ∠PRQ

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculate ∠QPR

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sine Rule in △PQR

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find α

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sine Rule in △PRS

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relation for β

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Evaluate 4αβsinθ∘

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplify Expression

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the Interval