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10th Class Mathematics Introduction to Trigonometry Question Bank Trigonometry

  • question_answer
    If \[tan(x+y)=\sqrt{3}\] and \[\mathbf{tan}\left( \mathbf{x}-\mathbf{y} \right)=\frac{1}{\sqrt{3}},\]\[\angle \mathbf{x}+\angle \mathbf{y}<\mathbf{9}{{\mathbf{0}}^{{}^\circ }},\mathbf{x}\ge \mathbf{y},\] then \[\angle \mathbf{x}\]is

    A) \[{{90}^{{}^\circ }}\]

    B) \[{{30}^{{}^\circ }}\]

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

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

    Correct Answer: C

    Solution :

    (c): \[tan(x+y)=\sqrt{3}=tan{{60}^{{}^\circ }}\] \[\Rightarrow x+y={{60}^{{}^\circ }}\]                        ....(i) \[tan\left( x-y \right)=\frac{1}{\sqrt{3}}=tan{{30}^{{}^\circ }}\] \[\Rightarrow x-y={{30}^{{}^\circ }}\]             .....(ii) \[\therefore x+y+x-y={{60}^{{}^\circ }}+{{30}^{{}^\circ }}\] \[\Rightarrow 2x={{90}^{{}^\circ }}\] \[\Rightarrow x=\frac{{{90}^{{}^\circ }}}{2}={{45}^{{}^\circ }}\]


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