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KVPY Sample Paper KVPY Stream-SX Model Paper-13

  • question_answer
    A triangle has base 10 cm long and the base angles of \[50{}^\circ \]and \[70{}^\circ .\] If the perimeter of the triangle is\[x+y\,\cos z{}^\circ ,\]where \[z\in (0,90{}^\circ ),\]then the value of \[x+y+z\]equals to

    A) 60

    B) 55      

    C) 50

    D) 40

    Correct Answer: D

    Solution :

    In \[\Delta ABC\]
    \[\frac{10}{\sin 60{}^\circ }=\frac{b}{\sin 50{}^\circ }=\frac{c}{\sin 70{}^\circ }=K\]
    \[\frac{20}{\sqrt{3}}=\frac{b}{\sin 50{}^\circ }=\frac{c}{\sin 70{}^\circ }=K\]
    \[b=\frac{20}{\sqrt{3}}\sin 50{}^\circ \]
    \[c=\frac{20}{\sqrt{3}}\sin 70{}^\circ \]
    \[\Rightarrow \]\[b+c=\frac{20}{\sqrt{3}}(\sin 50{}^\circ +\sin 70{}^\circ )\]
    \[=\frac{40}{\sqrt{3}}\sin 60{}^\circ \cos 10{}^\circ \]
    \[b+c=20\cos 10{}^\circ \]
    Perimeter of \[\Delta ABC\]
                \[=a+b+c=x+y\cos z{}^\circ \]
                \[=10+20\cos 10{}^\circ =x+y\cos z{}^\circ \]
                \[\therefore \]\[x=10,\]\[y=20,\]\[z=10\]
                \[\therefore \]\[x+y+z=10+20+10=40\]


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