Question

In a triangle, the lengths of two larger sides are $10cm$and $9cm$. If the angles of the triangle are in AP, then the length of the third side is:

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

Answer 1

The correct option is D

$5\pm \sqrt{6}$

Step 1: Given information and assumptions

Let the three angles be: $x-d,x,x+d$ .

Sum of all the interior angles is $180°$.

$x-d,x,x+d$ are in AP.

The two larger sides are $10cm$and $9cm$.

Step 2: Finding the angles

$$\begin{gathered} \Rightarrow x-d+x+x+d=180° \\ \Rightarrow 3x=180° \\ \therefore x=60°,x-d=30°,x+d=90° \end{gathered}$$

Step 3: Finding the third side

The triangle is a right-angled triangle.

The third side is the smallest side and is therefore not a hypotenuse.

$$\begin{gathered} \Rightarrow \cos \left(B\right)=\frac{a^2+c^2-b^2}{2ac} \\ \Rightarrow \cos \left(60\right)=\frac{{10}^2+c^2-9^2}{20c} \\ \Rightarrow \frac{1}{2}=\frac{{10}^2+c^2-9^2}{20c} \\ \Rightarrow {10}^2+c^2-9^2=10c \\ \Rightarrow c^2-10c\_+19=0 \\ \Rightarrow c=\frac{(10\pm \sqrt{100-76})}{2} \\ \Rightarrow c=\frac{10\pm \sqrt{24}}{2} \\ \Rightarrow c=5\pm \sqrt{6} \end{gathered}$$

Hence, Option ( D) is correct.