CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    If\[\sin \theta =\sin 15{}^\circ +\sin 45{}^\circ ,\]where\[0{}^\circ <\theta <90{}^\circ ,\] then\[\theta \]is equal to

    A)  \[45{}^\circ \]                                  

    B)  \[54{}^\circ \]

    C)  \[60{}^\circ \]                  

    D)         \[72{}^\circ \]

    E)  \[75{}^\circ \]

    Correct Answer: E

    Solution :

    Given, \[sin\theta =sin\text{ }15{}^\circ +sin\text{ }45{}^\circ \] \[=2\sin 30{}^\circ \cos 15{}^\circ =2\times \frac{1}{2}\times \cos (90{}^\circ -75{}^\circ )\] \[\Rightarrow \]\[\sin \theta =\sin 75{}^\circ \Rightarrow \theta =75{}^\circ \]


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