A) \[{{c}^{2}}-3c-7=0\]
B) \[{{c}^{2}}+3c+7=0\]
C) \[{{c}^{2}}-3c+7=0\]
D) \[{{c}^{2}}+3c-7=0\]
Correct Answer: A
Solution :
\[\cos A=\frac{{{b}^{2}}+{{c}^{2}}-{{\alpha }^{2}}}{2bc}\] Given, \[a=4,\,b=3\] and \[\angle A={{60}^{o}}\] \[\therefore \] \[\cos {{60}^{o}}=\frac{{{c}^{2}}+9-16}{2\times 3\times c}\] \[\Rightarrow \] \[\frac{1}{2}=\frac{{{c}^{2}}-7}{2\times 3c}\] \[\Rightarrow \] \[{{c}^{2}}-7=3c\] \[\Rightarrow \] \[{{c}^{2}}-3c-7=0\] Thus, c is the root of the above equation.
You need to login to perform this action.
You will be redirected in
3 sec
