J & K CET Engineering J and K - CET Engineering Solved Paper-2003

  • question_answer
    In a triangle ABC, \[a=5,\] \[b=4\] and \[\cos (A+B)=\frac{31}{32}\] In this triangle, c is equal to

    A)  \[\sqrt{6}\]           

    B)  \[36\]

    C)  \[6\]           

    D)  None of these

    Correct Answer: D

    Solution :

    Given that, \[a=5,\text{ }b=4\]and \[\cos \,(A+B)=\frac{31}{32}\] \[\Rightarrow \] \[\cos \,(\pi -C)=\frac{31}{32}\] \[\Rightarrow \] \[-\cos \,C=-\frac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}=\frac{31}{32}\] \[\Rightarrow \] \[\frac{{{(5)}^{2}}+{{(4)}^{2}}-{{c}^{2}}}{2\times 5\times 4}=-\frac{31}{32}\] \[\Rightarrow \] \[41-{{c}^{2}}=-\frac{155}{4}\] \[\Rightarrow \] \[{{c}^{2}}=41+\frac{155}{4}\] \[\Rightarrow \] \[{{c}^{2}}=\frac{319}{4}\] \[\Rightarrow \] \[c=\sqrt{\frac{319}{2}}\]


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