3}\] \[\Delta =\FRAC{1}{2}ab\sin C\Rightarrow ab=20\sqrt{3}\frac{2}{\sqrt{3}}=40\] ?.. (i) Also \[a+b+c=20\]or \[a+b=(20-c)\] Now, \[\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2AB}=\frac{1}{2}\] Þ \[{{a}^{2}}+{{b}^{2}}-{{c}^{2}}=ab\] Þ \[{{(a+b)}^{2}}-{{c}^{2}}=ab+2ab=3ab\] Þ \[{{(20-c)}^{2}}-{{c}^{2}}=3(40)\] Þ \[-40c+400=120\Rightarrow c=7\].

"> 3}\] \[\Delta =\FRAC{1}{2}ab\sin C\Rightarrow ab=20\sqrt{3}\frac{2}{\sqrt{3}}=40\] ?.. (i) Also \[a+b+c=20\]or \[a+b=(20-c)\] Now, \[\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2AB}=\frac{1}{2}\] Þ \[{{a}^{2}}+{{b}^{2}}-{{c}^{2}}=ab\] Þ \[{{(a+b)}^{2}}-{{c}^{2}}=ab+2ab=3ab\] Þ \[{{(20-c)}^{2}}-{{c}^{2}}=3(40)\] Þ \[-40c+400=120\Rightarrow c=7\].

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Area of the triangle is \[10\sqrt{3}\]sq. cm, angle \[C={{60}^{o}}\]and its perimeter is 20 cm, then side c will be

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 11 months ago

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\[\Delta =10\sqrt{3}\] \[\Delta =\FRAC{1}{2}ab\sin C\Rightarrow ab=20\sqrt{3}\frac{2}{\sqrt{3}}=40\] ?.. (i) Also \[a+b+c=20\]or \[a+b=(20-c)\] Now, \[\cos C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2AB}=\frac{1}{2}\] Þ \[{{a}^{2}}+{{b}^{2}}-{{c}^{2}}=ab\] Þ \[{{(a+b)}^{2}}-{{c}^{2}}=ab+2ab=3ab\] Þ \[{{(20-c)}^{2}}-{{c}^{2}}=3(40)\] Þ \[-40c+400=120\Rightarrow c=7\].

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Posted on 19 Aug 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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