If in a ΔABC, the altitudes from the vertices A, B, C on opposite sides are in H.P, then sin A, sin

Q: Wait, why do we use the area formula? Can't we just use the definition of altitude directly?

The altitude is a geometric segment, but to perform algebra, we need a bridge. The area formula $\Delta = \frac{1}{2}ah$ is that bridge. It links the altitude to the side length, allowing us to convert a geometric condition into an algebraic one.

Q: Why does the H.P. condition turn into an A.P. condition? Is there a shortcut?

H.P. is defined by the reciprocals being in A.P. There is no simpler way! It is the defining characteristic of Harmonic Progression. Whenever you see H.P., immediately think of the reciprocals.

Q: Can I use the Sine Rule on the altitudes directly?

No, the Sine Rule relates sides and angles, not altitudes. You must first convert the altitudes into sides using the area formula, and *then* use the Sine Rule to transition to the angles.

Question

If in a $\Delta ABC$, the altitudes from the vertices $A, B, C$ on opposite sides are in H.P, then $\sin A, \sin B, \sin C$ are in

Options:

(A) G.P.

(B) A. P.

(C) A.P-G.P

(D) H.P

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The area of a triangle $\Delta$ is defined by the fundamental relation:
$$\Delta = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$

This allows us to express the altitudes $h_1, h_2, h_3$ corresponding to sides $a, b, c$ as:
$$\Delta = \frac{1}{2} a h_1 = \frac{1}{2} b h_2 = \frac{1}{2} c h_3$$

By rearranging these equations, we obtain the expressions for the altitudes in terms of the area and side lengths:
$$h_1 = \frac{2\Delta}{a}, \quad h_2 = \frac{2\Delta}{b}, \quad h_3 = \frac{2\Delta}{c}$$

### The Harmonic Trap

We are given that the altitudes $h_1, h_2, h_3$ are in **Harmonic Progression (H.P.)**. By definition, if these terms are in H.P., their reciprocals must be in **Arithmetic Progression (A.P.)**:
$$\frac{1}{h_1}, \frac{1}{h_2}, \frac{1}{h_3} \in 	ext{A.P.}$$

Substituting our expressions for the altitudes into this sequence, we get:
$$\frac{a}{2\Delta}, \frac{b}{2\Delta}, \frac{c}{2\Delta} \in 	ext{A.P.}$$

Since $2\Delta$ is a constant for a given triangle, we can multiply the entire sequence by $2\Delta$ without affecting the progression. Consequently, the side lengths $a, b, c$ must be in **A.P.**

### The Sine Rule Bridge

To relate these sides to the angles of the triangle, we utilize the **Sine Rule**:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$

From this, we can express the sides as $a = 2R \sin A$, $b = 2R \sin B$, and $c = 2R \sin C$. Substituting these into our established A.P. sequence, we have:
$$2R \sin A, 2R \sin B, 2R \sin C \in 	ext{A.P.}$$

Dividing by the constant $2R$, we arrive at the final conclusion. The values **$\sin A, \sin B, \sin C$ are in Arithmetic Progression.**

Animated Solution for Mathematics - Trigonometry: If in a $Δ A B C$ , the altitudes from the vertices $A, B, C$ on opposite sides are in H.P, then $sin A, sin B, sin C$ are in

Select Answer:

Visualized Solution

Visualizing the Altitudes

Area of Triangle Relation

Expressing Altitudes

Defining the H.P. Condition

Substituting Altitudes into A.P.

Simplifying the Progression

Applying the Sine Rule

Expressing Sides via Sine Rule

Final Conclusion

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Harmonic Trap

The Sine Rule Bridge

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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 ΔABC, the altitudes from the vertices A,B,C on opposite sides are in H.P, then sinA,sinB,sinC are in