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Question: Answered & Verified by Expert
In a triangle $\mathrm{ABC}$ if the angles $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are in $\mathrm{AP}$, then which one of the following is correct?
MathematicsProperties of TrianglesNDANDA 2012 (Phase 1)
Options:
  • A $\quad \mathrm{c}=\mathrm{a}+\mathrm{b}$
  • B $c^{2}=a^{2}+b^{2}-a b$
  • C $a^{2}=b^{2}+c^{2}-b c$
  • D $\quad b^{2}=a^{2}+c^{2}-a c$
Solution:
2868 Upvotes Verified Answer
The correct answer is: $\quad b^{2}=a^{2}+c^{2}-a c$
Since $\mathrm{A}, \mathrm{B}, \mathrm{C}$ are in A.P. $\therefore 2 \mathrm{~B}=\mathrm{A}+\mathrm{C}$
Also, $\mathrm{A}+\mathrm{B}+\mathrm{C}=180^{\circ} \Rightarrow 2 \mathrm{~B}+\mathrm{B}=180^{\circ}$
$\Rightarrow 3 \mathrm{~B}=180^{\circ} \Rightarrow \mathrm{B}=60^{\circ}$
Now, we know
$\cos B=\frac{a^{2}+c^{2}-b^{2}}{2 a c}$
$\Rightarrow \cos 60^{\circ}=\frac{\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2}}{2 \mathrm{ac}}$
$\Rightarrow \frac{1}{2}=\frac{\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2}}{2 \mathrm{ac}} \Rightarrow \mathrm{ac}=\mathrm{a}^{2}+\mathrm{c}^{2}-\mathrm{b}^{2}$
$\Rightarrow b^{2}=a^{2}+c^{2}-a c$

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