CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2005

  • question_answer
    If   the   sides   of   the   triangle   are \[p,q,\sqrt{{{p}^{2}}+{{q}^{2}}+pq},\]then the greatest angle is:

    A)  \[\frac{\pi }{2}\]             

    B)                         \[\frac{5\pi }{4}\]

    C)  \[\frac{2\pi }{3}\]                           

    D)         \[\frac{7\pi }{4}\]

    E)  \[\frac{5\pi }{3}\]

    Correct Answer: C

    Solution :

    The longest side of triangle is\[\sqrt{{{p}^{2}}+{{q}^{2}}+pq}\]. Greatest angle will be opposite to longest side. Let\[\theta \]be greatest angle, then \[\cos \theta =\frac{{{p}^{2}}+{{q}^{2}}-{{p}^{2}}-{{q}^{2}}-pq}{2pq}=-\frac{1}{2}\] \[\Rightarrow \]               \[\cos \theta =\cos \frac{2\pi }{3}\] \[\Rightarrow \]               \[\theta =\frac{2\pi }{3}\]


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