Question

In a triangle ABC, $m\angle A$, $m\angle B$, $m\angle C$ are in A.P. and lengths of two larger sides are 10 units, 9 units respectively, then the length (in units) of the third side is

• A. $5 + \sqrt{6}$
• B. $\sqrt{5} - 1$
• C. $\sqrt{6} + 1$
• D. $\sqrt{5} + 1$

Hint:

For angles in A.P. in a triangle, the middle angle is always $60^\circ$. Use the Law of Cosines to find the unknown side. Remember to check both possible solutions from the quadratic equation if applicable and ensure they fit the context of the problem.

Answer 1

The Correct Option is A

Step 1: Determine the value of angle $B$. Given that $m\angle A$, $m\angle B$, $m\angle C$ are in Arithmetic Progression (A.P.). By the property of A.P., the middle term is the average of the other two: \[ 2m\angle B = m\angle A + m\angle C \quad \text{.... (1)} \] The sum of angles in a triangle is $180^\circ$: \[ m\angle A + m\angle B + m\angle C = 180^\circ \quad \text{.... (2)} \] Substitute (1) into (2): \[ (2m\angle B) + m\angle B = 180^\circ \] \[ 3m\angle B = 180^\circ \] \[ m\angle B = \frac{180^\circ}{3} = 60^\circ \]
Step 2: Assign the lengths of the given sides. Let the sides opposite to angles $A, B, C$ be $a, b, c$ respectively. We are given that the lengths of two larger sides are 10 units and 9 units. Let $a = 10$ and $b = 9$. We need to find the length of the third side, $c$.
Step 3: Apply the Law of Cosines. Using the Law of Cosines for side $b$: \[ b^2 = a^2 + c^2 - 2ac \cos B \] Substitute the known values $b=9$, $a=10$, and $B=60^\circ$: \[ 9^2 = 10^2 + c^2 - 2(10)(c) \cos 60^\circ \] We know that $\cos 60^\circ = \frac{1}{2}$: \[ 81 = 100 + c^2 - 20c \left(\frac{1}{2}\right) \] \[ 81 = 100 + c^2 - 10c \]
Step 4: Solve the quadratic equation for $c$. Rearrange the equation into a standard quadratic form $Ac^2 + Bc + C = 0$: \[ c^2 - 10c + 100 - 81 = 0 \] \[ c^2 - 10c + 19 = 0 \] Use the quadratic formula $c = \frac{-B \pm \sqrt{B^2 - 4AC{2A}$: \[ c = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(19){2(1)} \] \[ c = \frac{10 \pm \sqrt{100 - 76{2} \] \[ c = \frac{10 \pm \sqrt{24{2} \] Simplify $\sqrt{24} = \sqrt{4 \times 6} = 2\sqrt{6}$: \[ c = \frac{10 \pm 2\sqrt{6{2} \] \[ c = 5 \pm \sqrt{6} \]
Step 5: Choose the valid length for side $c$. Both $5 + \sqrt{6}$ and $5 - \sqrt{6}$ are positive values ($5 + \sqrt{6} \approx 5 + 2.45 = 7.45$, and $5 - \sqrt{6} \approx 5 - 2.45 = 2.55$). The sides are $a=10$, $b=9$, and $c$. If $c = 5+\sqrt{6} \approx 7.45$, then the sides are $10, 9, 7.45$. In this case, 10 and 9 are indeed two larger sides. This option matches one of the given choices. If $c = 5-\sqrt{6} \approx 2.55$, then the sides are $10, 9, 2.55$. In this case, 10 and 9 are also two larger sides. However, this is not among the options. Given the options, $5+\sqrt{6}$ is the correct choice.