If in a triangle PQR, sin P, sin Q, sin R are in A.P., then (a) the altitudes are in A.P. (b) the al

Q: Wait, why can't we just assume the triangle is equilateral to solve this?

That is a clever shortcut for a quick check, but in JEE Advanced, we must prove the result for the general case. Assuming an equilateral triangle only proves the property holds for one specific case, not for all triangles satisfying the condition. The algebraic proof we derived is universal and holds for any triangle where the sines are in A.P.

Q: Why do we use the area formula? Is there a faster way?

The area formula is the most elegant bridge between the side lengths and the altitudes. It directly links the geometric properties (altitudes) to the side lengths. While you could potentially use other trigonometric identities, this method is the most robust and least prone to calculation errors.

Q: How do I remember the relationship between A.P. and H.P.?

Think of it as a reciprocal dance! If a sequence $a, b, c$ is in Arithmetic Progression, then their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in Harmonic Progression. It is a fundamental property of sequences that you should keep in your toolkit for all competitive exams.

Question

If in a triangle $PQR, \sin P, \sin Q, \sin R$ are in A.P., then

Options:

(A) the altitudes are in A.P.

(B) the altitudes are in H.P.

(C) the medians are in G.P.

(D) the medians are in A.P.

Sigma Percentile · Accepted Answer

### The Harmonic Dance of Geometry

Welcome, my dear student. Today, we are not just solving a problem; we are uncovering a beautiful, hidden symmetry within the geometry of a triangle.

Often, when we look at a triangle $PQR$, we see just three sides and three angles. But to a mathematician, a triangle is a system of interconnected ratios. Let us embark on this journey to understand why, when the sines of the angles are in Arithmetic Progression (A.P.), the altitudes must follow the rhythm of a Harmonic Progression (H.P.).

### Phase 1: The Sine Rule Bridge

Imagine you are a surveyor standing at the vertices of triangle $PQR$. You have the sides $p, q, r$ opposite to the angles $P, Q, R$.

The most powerful tool in your arsenal to connect these sides to the angles is the Sine Rule:

$$
\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = k
$$

Here, $k$ is a constant for our specific triangle. By rearranging this, we can express each side length purely in terms of its opposite angle and this constant $k$:

$$
p = k \sin P, \quad q = k \sin Q, \quad r = k \sin R
$$

This is our first major step. We have successfully translated the geometry of side lengths into the language of trigonometry.

### Phase 2: The Area Connection

Now, let us consider the heartbeat of the triangle: its area, $\Delta$. We know the fundamental formula for the area of any triangle is $\Delta = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$.

Since a triangle has three sides, we can calculate its area in three different ways, depending on which side we choose as the base:

$$
\Delta = \frac{1}{2} p h_1 = \frac{1}{2} q h_2 = \frac{1}{2} r h_3
$$

Here, $h_1, h_2, h_3$ are the altitudes dropped onto sides $p, q, r$ respectively. Our goal is to understand the behavior of these altitudes. Let us isolate them:

$$
h_1 = \frac{2\Delta}{p}, \quad h_2 = \frac{2\Delta}{q}, \quad h_3 = \frac{2\Delta}{r}
$$

### Phase 3: The Harmonic Revelation

Now, let us bring our two worlds together. Substitute the expressions for $p, q, r$ from our Sine Rule into these altitude equations:

$$
h_1 = \frac{2\Delta}{k \sin P}, \quad h_2 = \frac{2\Delta}{k \sin Q}, \quad h_3 = \frac{2\Delta}{k \sin R}
$$

Look closely at these expressions. The term $\frac{2\Delta}{k}$ is a constant for our triangle. This reveals a profound truth: the altitudes are inversely proportional to the sines of their opposite angles:

$$
h_1 \propto \frac{1}{\sin P}, \quad h_2 \propto \frac{1}{\sin Q}, \quad h_3 \propto \frac{1}{\sin R}
$$

The problem states that $\sin P, \sin Q, \sin R$ are in A.P. By definition, if a sequence of numbers is in A.P., their reciprocals are in H.P.

Since our altitudes $h_1, h_2, h_3$ are proportional to the reciprocals of the sines, they must also follow the same progression. Therefore, the altitudes $h_1, h_2, h_3$ are in **H.P.**

### Conclusion

Isn't it magnificent? We started with a simple condition about angles and arrived at a structural property of the triangle's altitudes.

This is the beauty of mathematics—it reveals the hidden order beneath the surface. Keep practicing, keep visualizing, and never stop asking 'why'. You are well on your way to mastering these concepts.

Animated Solution for Mathematics - Trigonometry: If in a triangle $P QR, sin P, sin Q, sin R$ are in A.P., then

Select Answer:

Visualized Solution

Visualizing Triangle $P QR$

Introducing the Altitudes

The Sine Rule

Expressing Sides using Sine

Area of the Triangle

Area with Different Bases

Isolating the Altitudes

Substituting Sine Values

Altitudes and Sine Proportionality

The Given A.P. Condition

Final Conclusion

The Sigma Insight: Properties of Triangles

The Harmonic Dance of Geometry

Phase 1

Phase 2

Phase 3

Conclusion

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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; } If in a triangle PQR,sinP,sinQ,sinR are in A.P., then