Question

In a triangle, the lengths of the two largest sides are \(10\) cm. and \(9\) cm. If the angles of the triangle are in arithmatic progression (A. P.), then the length of the third side (in cm.) is...

• A. \(3\pm \sqrt{6}\)
• B. \(4\pm \sqrt{6}\)
• C. \(5\pm \sqrt{6}\)
• D. \(6\pm \sqrt{6}\)

Hint:

Angles in A.P. force the middle angle to be 60 degrees, and then apply the cosine rule to the middle side.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
If the angles are in A.P., the middle angle satisfies \(2B=A+C\), and with \(A+B+C=180^{\circ}\) we get \(B=60^{\circ}\).

Step 2: Identify the Sides
The two largest sides are 10 and 9, so the middle side is 9 and the smallest side is the unknown x.
The middle side 9 is opposite the \(60^{\circ}\) angle.

Step 3: Cosine Rule
\(9^{2}=10^{2}+x^{2}-2(10)(x)\cos 60^{\circ}\)
\(81=100+x^{2}-10x\)
\(x^{2}-10x+19=0\).

Step 4: Solve
\(x=\dfrac{10\pm\sqrt{100-76}}{2}=5\pm\sqrt{6}\).
Both values are positive and less than 9, so both are valid.

Final Answer:
The third side is \(5\pm\sqrt{6}\) cm, option (C). \[ \boxed{\text{(C) }5\pm \sqrt{6}} \]