Question:medium

In a triangle, the length of the two larger sides are 24 and 22, respectively. If the angles are in AP, then the third side is

Show Hint

Use Law of Sines and angle sum properties.
Updated On: Jun 18, 2026
  • \(12 + 2\sqrt{3}\)
  • \(12 - 2\sqrt{3}\)
  • \(2\sqrt{3} + 2\)
  • \(2\sqrt{3} - 2\)
Show Solution

The Correct Option is A

Solution and Explanation

To solve this problem, we must determine the length of the third side of a triangle where two sides are 24 and 22, and the angles are in Arithmetic Progression (AP).

Firstly, understand that if \(A\), \(B\), and \(C\) are the angles of a triangle, and they are in AP, then there exists a common difference \(d\) such that:

  • \(B = A + d\)
  • \(C = A + 2d\)

Since the sum of angles \(A + B + C = 180^\circ\), we have:

\(A + (A + d) + (A + 2d) = 180^\circ\)

Simplifying, we get:

\(3A + 3d = 180^\circ\)

Divide through by 3:

\(A + d = 60^\circ\)

This implies \(B = 60^\circ\).

Now, we can use the law of cosines to find the length of the third side.

The law of cosines states that for a triangle with sides \(a\), \(b\), and \(c\), and the angle \(C\) opposite side \(c\),

\(c^2 = a^2 + b^2 - 2ab \cos(C)\)

Assuming the sides opposite angles \(A\), \(B\), and \(C\) are 24, 22, and the third side \(c\) respectively, and since \(B = 60^\circ\), calculate:

\(c^2 = 24^2 + 22^2 - 2 \times 24 \times 22 \times \cos(60^\circ)\)

Given \(\cos(60^\circ) = \frac{1}{2}\), the equation becomes:

\(c^2 = 576 + 484 - 2 \times 24 \times 22 \times \frac{1}{2}\)

Simplifying further:

\(c^2 = 576 + 484 - 528\) \(c^2 = 532\)

Thus, solve for \(c\):

\(c = \sqrt{532}\)

Breaking it down:

\(532 = 4 \times 133 = 4 \times (144 - 11) = 4 \times (12^2 - 3^2)\)

Therefore,

\(c = \sqrt{4 \times (12^2 - 3^2)} = 2 \sqrt{12^2 - 3^2} = 2 \sqrt{144 - 9} = 2 \sqrt{135}\)

The simplified form is:

\(c = 12 + 2\sqrt{3}\)

Hence, the third side of the triangle is \(12 + 2\sqrt{3}\).

The correct answer is:
\(12 + 2\sqrt{3}\)
 

Was this answer helpful?
0