The angles $A, B, C$ of a triangle $A B C$ are in AP. If $A B=6, B C=7$, then $A C=$

1. $\sqrt{40}$
2. $\sqrt{41}$
3. $\sqrt{43}$
4. 6

Solution

$\begin{aligned} & \text { (c) Given In } \triangle A B C \text {, } \\ & \angle A, \angle B, \angle C \text { are in AP } \\ & \Rightarrow \\ & \text { In } \triangle A B C,\end{aligned}$
$ \begin{aligned} \angle A+\angle B+\angle C & =180^{\circ} \\ 2 \angle B+\angle B & =180^{\circ} \\ \Rightarrow \quad \angle B & =60^{\circ} \end{aligned} $ From cosine rule, $ \begin{aligned} \cos B & =\frac{A B^2+B C^2-A C^2}{2 A B \times B C} \\ \cos 60^{\circ} & =\frac{36+49-x^2}{2 \times 6 \times 7} \\ \frac{1}{2} & =\frac{85-x^2}{84} \\ 41 & =85-x^2 \end{aligned} $ $ \begin{aligned} x^2 & =85-42 \\ x^2 & =43 \\ x & =\sqrt{43} \end{aligned} $ Hence, option (3) is correct

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

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