(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. (stackrelfrown{POQ} = 160°), (stackrelfrown{QSR} = 45°) and (stackrelfrown{PQS} = 40°). Calculate, (i) < QPS ; (ii) < RQS.
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 = 2sqrt{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°)