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(a) Simplify : \(\sqrt{1001_{two}}\), leaving your answer in base two. (b) In the ...



(a) Simplify : \(\sqrt{1001_{two}}\), leaving your answer in base two.
(b)
In the diagram, O is the centre of the circle radius x. /PQ/ = z, /OK/ = y and < OKP = 90°. Find the value of z in terms of x and y.
(c)
In the diagram, P, Q, R and S are points of the circle centre O. \(\stackrel\frown{POQ} = 160°\), \(\stackrel\frown{QSR} = 45°\) and \(\stackrel\frown{PQS} = 40°\). Calculate, (i) < QPS ; (ii) < RQS.
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    Correct Answer: Option n
    Explanation:
    (a) \(1001_{2} = 1 \times 2^{3} + 0 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0}\)
    = \(8 + 0 + 0 + 1\)
    = 9
    \(\sqrt{9} = 3_{10}\)
    2 3
    2 1 r 1
    0 r 1



    \(\therefore \sqrt{1001_{2}} = 11_{2}\)
    (b) In \(\Delta QOP, OQ = OP\)
    \(\therefore QK = KP\)
    \(\Delta QOK = \Delta POK\) (right- angled triangle)
    In \(\Delta POK\),
    \(OP^{2} = PK^{2} + OK^{2}\)
    \(x^{2} = y^{2} + (\frac{z}{2})^{2}\)
    \(\frac{z^{2}}{4} = x^{2} - y^{2}\)
    \(z^{2} = 4x^{2} - 4y^{2}\)
    \(z = \sqrt{4(x^{2} - y^{2})}\)
    \(z = 2\sqrt{x^{2} - y^{2}}\)
    (c)(i) \(< PSQ = \frac{1}{2}(< POQ) = \frac{1}{2}(160°)\)
    = 80°
    In \(\Delta PSQ\),
    \(< PQS + < QPS + < PSQ = 180°\)
    \(40° + < QPS + 80° = 180°\)
    \(< QPS = 180° - 120° = 60°\)
    (ii) \(< PSR = < PSQ + < QSR\)
    = \(80° + 45°\)
    = \(125°\)
    \(< PQS = < PQO + < RQS \)
    = \(40° + < RQS\)
    \(< PSR + < PQS = 180°\)
    \(\therefore 125° + 40° + < RQS = 180°\)
    \(< RQS = 180° - 165°\)
    = \(15°\)

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