C are in A. P. then angle\[B={{60}^{o}},\] \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\], \[\left\{ \begin{align}   & \text{since }A+B+C={{180}^{o}}\,\,\text{and} \\  & \text{        }A+C=2B\RIGHTARROW B={{60}^{o}} \\ \END{align} \right\}\] Þ \[\frac{1}{2}=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\Rightarrow {{a}^{2}}+{{c}^{2}}-{{b}^{2}}=ac\] Þ \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-ac\].

"> C are in A. P. then angle\[B={{60}^{o}},\] \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\], \[\left\{ \begin{align}   & \text{since }A+B+C={{180}^{o}}\,\,\text{and} \\  & \text{        }A+C=2B\RIGHTARROW B={{60}^{o}} \\ \END{align} \right\}\] Þ \[\frac{1}{2}=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\Rightarrow {{a}^{2}}+{{c}^{2}}-{{b}^{2}}=ac\] Þ \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-ac\].

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If the angles of a triangle \[ABC\]be in A.P., then

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 10 months ago

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A, B, C are in A. P. then angle\[B={{60}^{o}},\] \[\cos B=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\], \[\left\{ \begin{align}   & \text{since }A+B+C={{180}^{o}}\,\,\text{and} \\  & \text{        }A+C=2B\RIGHTARROW B={{60}^{o}} \\ \END{align} \right\}\] Þ \[\frac{1}{2}=\frac{{{a}^{2}}+{{c}^{2}}-{{b}^{2}}}{2ac}\Rightarrow {{a}^{2}}+{{c}^{2}}-{{b}^{2}}=ac\] Þ \[{{b}^{2}}={{a}^{2}}+{{c}^{2}}-ac\].

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Posted on 11 Sep 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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