SIDE is\[\SQRT{{{p}^{2}}+pq+{{Q}^{2}}}\]. If largest angle is \[\THETA \], then \[\cos \theta =\frac{{{p}^{2}}+{{q}^{2}}-{{p}^{2}}-pq-{{q}^{2}}}{2pq}=-\frac{1}{2}=\cos \left( \frac{2\pi }{3} \right)\] Þ \[\theta =\frac{2\pi }{3}\].

"> SIDE is\[\SQRT{{{p}^{2}}+pq+{{Q}^{2}}}\]. If largest angle is \[\THETA \], then \[\cos \theta =\frac{{{p}^{2}}+{{q}^{2}}-{{p}^{2}}-pq-{{q}^{2}}}{2pq}=-\frac{1}{2}=\cos \left( \frac{2\pi }{3} \right)\] Þ \[\theta =\frac{2\pi }{3}\].

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If the sides of a triangle are p,q and\[\sqrt{{{p}^{2}}+pq+{{q}^{2}}}\], then the biggest angle is   [Kerala (Engg.) 2005]

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

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Largest SIDE is\[\SQRT{{{p}^{2}}+pq+{{Q}^{2}}}\]. If largest angle is \[\THETA \], then \[\cos \theta =\frac{{{p}^{2}}+{{q}^{2}}-{{p}^{2}}-pq-{{q}^{2}}}{2pq}=-\frac{1}{2}=\cos \left( \frac{2\pi }{3} \right)\] Þ \[\theta =\frac{2\pi }{3}\].

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Posted on 15 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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