A) \[C=A+B\]
B) \[C=2B\]
C) \[C=2A\]
D) \[C=3A\]
Correct Answer: C
Solution :
Given, in \[\Delta \,ABC,\,\,a=8\,cm,\,\,b=10cm,\,c=12\,cm\] \[\cos \,C=\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}=\frac{64+100-144}{2.8.10}\] \[=\frac{20}{2.8.10}=\frac{1}{8}\] ?..(i) \[\cos \,A=\frac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}\] \[=\frac{100+144-64}{2.10.12}=\frac{180}{2.10.12}=\frac{3}{4}\] \[\cos \,2A=2\,{{\cos }^{2}}A-1=2.\,{{\left( \frac{3}{4} \right)}^{2}}-1\] \[=\frac{2.9}{16}-1=\frac{8}{9}-1=\frac{1}{8}\] ?.(ii) From Eqs. (i) and (ii), we get \[\cos \,C=\cos \,2A\] \[\Rightarrow \] \[C=2A\]
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