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Question

For the next two (02) items that follow:

Consider a triangle ABC satisfying

\(2{\rm{a}}{\sin ^2}\left( {\frac{{\rm{C}}}{2}} \right) + 2{\rm{c}}{\sin ^2}\left( {\frac{{\rm{A}}}{2}} \right) = 2{\rm{a}} + 2{\rm{c}} - 3{\rm{b}}\)

The sides of the triangle are in

This question was previously asked in
NDA II 2015 GAT Previous Year Paper (16-Dec-2015)
The correct answer is

A.P.

This problem asks us to determine the relationship between the sides of a triangle ABC given a specific trigonometric equation involving its sides and angles. The given equation is: \(2{\rm{a}}{\sin ^2}\left( {\frac{{\rm{C}}}{2}} \right) + 2{\rm{c}}{\sin ^2}\left( {\frac{{\rm{A}}}{2}} \right) = 2{\rm{a}} + 2{\rm{c}} - 3{\rm{b}}\).

Analyzing the Triangle Equation

We are given an equation that connects the sides (a, b, c) and angles (A, C) of triangle ABC. To find the relationship between the sides (whether they are in A.P., G.P., or H.P.), we need to simplify this equation. We can use standard trigonometric identities and the Law of Cosines.

Applying Half-Angle Formulas

Recall the half-angle formula for sine squared: \({\sin ^2}\left( {\frac{x}{2}} \right) = \frac{{1 - \cos x}}{2}\). Applying this to the terms involving angles A and C:

  • \({\sin ^2}\left( {\frac{{\rm{C}}}{2}} \right) = \frac{{1 - \cos C}}{2}\)
  • \({\sin ^2}\left( {\frac{{\rm{A}}}{2}} \right) = \frac{{1 - \cos A}}{2}\)

Substitute these into the given equation:

\[2{\rm{a}}\left( {\frac{{1 - \cos C}}{2}} \right) + 2{\rm{c}}\left( {\frac{{1 - \cos A}}{2}} \right) = 2{\rm{a}} + 2{\rm{c}} - 3{\rm{b}}\]

Simplify the equation:

\[{\rm{a}}(1 - \cos C) + {\rm{c}}(1 - \cos A) = 2{\rm{a}} + 2{\rm{c}} - 3{\rm{b}}\]

\[{\rm{a}} - {\rm{a}}\cos C + {\rm{c}} - {\rm{c}}\cos A = 2{\rm{a}} + 2{\rm{c}} - 3{\rm{b}}\]

Rearrange the terms to isolate the \(3b\) term:

\[3{\rm{b}} = 2{\rm{a}} + 2{\rm{c}} - {\rm{a}} + {\rm{a}}\cos C - {\rm{c}} + {\rm{c}}\cos A\]

\[3{\rm{b}} = {\rm{a}} + {\rm{c}} + {\rm{a}}\cos C + {\rm{c}}\cos A\]

Using the Law of Cosines

Now, we use the Law of Cosines to express \(\cos A\) and \(\cos C\) in terms of the sides a, b, and c:

  • \({\cos A} = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\)
  • \({\cos C} = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\)

Substitute these expressions into the equation \(3{\rm{b}} = {\rm{a}} + {\rm{c}} + {\rm{a}}\cos C + {\rm{c}}\cos A\):

\[3{\rm{b}} = {\rm{a}} + {\rm{c}} + {\rm{a}}\left( {\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}} \right) + {\rm{c}}\left( {\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}} \right)\]

\[3{\rm{b}} = {\rm{a}} + {\rm{c}} + \frac{{{a^2} + {b^2} - {c^2}}}{{2b}} + \frac{{{b^2} + {c^2} - {\rm{a}}^2}}{{2b}}\]

Simplifying the Equation

To eliminate the denominators, multiply the entire equation by \(2b\):

\[3{\rm{b}}(2{\rm{b}}) = ( {\rm{a}} + {\rm{c}} )(2{\rm{b}}) + \left( {\frac{{{a^2} + {b^2} - {c^2}}}{{2b}}} \right)(2b) + \left( {\frac{{{b^2} + {c^2} - {\rm{a}}^2}}{{2b}}} \right)(2b)\]

\[6{{\rm{b}}^2} = 2{\rm{ab}} + 2{\rm{cb}} + {{\rm{a}}^2} + {{\rm{b}}^2} - {{\rm{c}}^2} + {{\rm{b}}^2} + {{\rm{c}}^2} - {{\rm{a}}^2}\]

Combine like terms on the right side. The \(a^2\) and \(-a^2\) terms cancel, and the \(c^2\) and \(-c^2\) terms cancel:

\[6{{\rm{b}}^2} = 2{\rm{ab}} + 2{\rm{cb}} + 2{{\rm{b}}^2}\]

Subtract \(2{{\rm{b}}^2}\) from both sides:

\[6{{\rm{b}}^2} - 2{{\rm{b}}^2} = 2{\rm{ab}} + 2{\rm{cb}}\]

\[4{{\rm{b}}^2} = 2{\rm{b}}({\rm{a}} + {\rm{c}})\]

Since b is a side of a triangle, \(b \neq 0\). We can divide both sides by \(2b\):

\[\frac{{4{{\rm{b}}^2}}}{{2{\rm{b}}}} = \frac{{2{\rm{b}}({\rm{a}} + {\rm{c}})}}{{2{\rm{b}}}}\]

\[2{\rm{b}} = {\rm{a}} + {\rm{c}}\]

Conclusion: Sides in Arithmetic Progression

The relationship \(2b = a + c\) is the defining condition for three numbers a, b, and c to be in Arithmetic Progression (A.P.). In an A.P., the middle term is the average of the first and third terms, or equivalently, the difference between consecutive terms is constant (\(b - a = c - b\), which simplifies to \(2b = a + c\)).

Thus, the sides a, b, and c of the triangle satisfy the condition for being in Arithmetic Progression.

Summary of Findings

Starting from the given trigonometric equation involving the sides and angles of triangle ABC, we used half-angle formulas and the Law of Cosines to transform the equation. Through algebraic simplification, we derived the relationship \(2b = a + c\), which proves that the sides of the triangle are in Arithmetic Progression (A.P.).

Revision Table: Progression Types

Understanding different types of progressions is key in sequence and series problems. Here's a quick review:

Progression Type Condition for terms \(x, y, z\) Description
Arithmetic Progression (A.P.) \(2y = x + z\) or \(y - x = z - y\) Each term after the first is obtained by adding a constant difference to the preceding term.
Geometric Progression (G.P.) \(y^2 = xz\) or \(\frac{y}{x} = \frac{z}{y}\) Each term after the first is obtained by multiplying the preceding term by a constant ratio.
Harmonic Progression (H.P.) \(\frac{2}{y} = \frac{1}{x} + \frac{1}{z}\) or \(\frac{1}{x}, \frac{1}{y}, \frac{1}{z}\) are in A.P. The reciprocals of the terms are in Arithmetic Progression.

Additional Information: Triangle Laws

Several fundamental laws govern the relationships between the sides and angles of a triangle. Two crucial ones used in this solution are:

  • Law of Sines: \(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\), where R is the circumradius of the triangle.
  • Law of Cosines: Relates a side of a triangle to the other two sides and the angle between them.
    • \(a^2 = b^2 + c^2 - 2bc \cos A\)
    • \(b^2 = a^2 + c^2 - 2ac \cos B\)
    • \(c^2 = a^2 + b^2 - 2ab \cos C\)
    From these, we can derive expressions for cosines of angles, e.g., \({\cos A} = \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\).
  • Half-Angle Formulas (in terms of sides): These can also be expressed directly using the sides and semi-perimeter \(s = \frac{a+b+c}{2}\).
    • \({\sin ^2}\left( {\frac{{\rm{A}}}{2}} \right) = \frac{{(s - b)(s - c)}}{{bc}}\)
    • \({\sin ^2}\left( {\frac{{\rm{C}}}{2}} \right) = \frac{{(s - a)(s - b)}}{{ab}}\)
    Using these formulas directly in the initial equation might also lead to the same result, potentially through a different algebraic path.
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Similar Questions

  1. In a triangle ABC, a = (1 + √3) cm, b = 2 cm and angle C = 60°, then the other two angles are

  2. Consider the following statements :

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Important Questions from Properties of Triangles

  1. In a triangle ABC, a = (1 + √3) cm, b = 2 cm and angle C = 60°, then the other two angles are

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