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9th Class Mathematics Triangles Question Bank Triangle

  • question_answer
    In the figure given below, an isosceles triangle ABC is right angled at B. D is a point inside the triangle ABC. P and Q are the point of the perpendiculars drawn from D on the side AB and AC respectively of \[\Delta \mathbf{ABC}\]. If \[AP=x\] cm, \[\mathbf{AQ=y}\] cm and \[\angle \mathbf{BAD}=\mathbf{1}{{\mathbf{5}}^{{}^\circ }}\]. Then, sin\[{{75}^{{}^\circ }}=\]?

    A)  \[\frac{2y}{\sqrt{3x}}\]            

    B)  \[\frac{x}{2y}\]

    C)  \[\frac{\sqrt{3x}}{2y}\]

    D)  \[\frac{2x}{\sqrt{3}y}\]

    Correct Answer: C

    Solution :

    (c): From \[\Delta AQD,\] \[\sin {{60}^{{}^\circ }}=\frac{AQ}{AD}\] \[\Rightarrow \]\[\frac{\sqrt{3}}{2}=\frac{y}{AD}\] \[\Rightarrow \]\[AD=\frac{2y}{\sqrt{3}}\] From \[\Delta APD,\] \[sin{{75}^{{}^\circ }}=\frac{AP}{AD}=\frac{x}{\frac{2y}{\sqrt{3}}}=\frac{\sqrt{3}x}{2y}\]              


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