In a triangle $\mathrm{ABC}, \mathrm{m} \angle \mathrm{A}, \mathrm{m} \angle \mathrm{B}, \mathrm{m} \angle…

In a triangle $\mathrm{ABC}, \mathrm{m} \angle \mathrm{A}, \mathrm{m} \angle \mathrm{B}, \mathrm{m} \angle \mathrm{C}$ 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
  1. $5+\sqrt{6}$
  2. $\sqrt{5}-1$
  3. $\sqrt{6}+1$
  4. $\sqrt{5}+1$

Solution

$\angle \mathrm{A}, \angle \mathrm{B}, \angle \mathrm{C}$ are in A.P. $\begin{aligned} & \Rightarrow 2 \mathrm{~B}=\mathrm{A}+\mathrm{C} \\ & \Rightarrow 3 \mathrm{~B}=\mathrm{A}+\mathrm{B}+\mathrm{C} \\ & \Rightarrow 3 \mathrm{~B}=180^{\circ} \\ & \Rightarrow \mathrm{B}=60^{\circ} \\ & \cos \mathrm{B}=\frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ca}} \end{aligned}$ $\Rightarrow \cos 60^{\circ}=\frac{\mathrm{c}^2+10^2-9^2}{2 \mathrm{c}(10)}$ $\ldots[$ Let $a=10, b=9]$ $\begin{aligned} & \Rightarrow \frac{1}{2}=\frac{\mathrm{c}^2+100-81}{20 \mathrm{c}} \\ & \Rightarrow 10 \mathrm{c}=\mathrm{c}^2+19 \\ & \Rightarrow \mathrm{c}^2-10 \mathrm{c}+19=0 \\ & \Rightarrow \mathrm{c}=\frac{10 \pm \sqrt{100-76}}{2} \\ & =\frac{10 \pm 2 \sqrt{6}}{2} \\ & =5 \pm \sqrt{6}\end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 2)

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