(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 < ...

Answer Details

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

First convert \(1001_{two}\) to base ten:

\[1001_{two}=1(2^3)+0(2^2)+0(2^1)+1(2^0)=8+0+0+1=9_{ten}.\]

Then \(\sqrt{9_{ten}}=3_{ten}\). Convert \(3\) back to base two:

\[3_{ten}=1(2^1)+1(2^0)=11_{two}.\]

Therefore \(\sqrt{1001_{two}}=\mathbf{11_{two}}\).

(b) Find \(z\) in terms of \(x\) and \(y\).

In the diagram, \(O\) is the centre, \(|OP|=x\) (a radius), \(|OK|=y\) and \(\angle OKP=90^\circ\). Since \(OK\) is drawn from the centre perpendicular to the chord \(PQ\), it bisects the chord, so \(K\) is the mid-point of \(PQ\):

\[|KP|=\tfrac{1}{2}|PQ|=\tfrac{z}{2}.\]

Applying Pythagoras' theorem to right-angled triangle \(OKP\):

\[|OP|^2=|OK|^2+|KP|^2\]\[x^2=y^2+\left(\frac{z}{2}\right)^2\]\[\left(\frac{z}{2}\right)^2=x^2-y^2\]\[\frac{z}{2}=\sqrt{x^2-y^2}\]

Therefore \(\displaystyle z=2\sqrt{x^{2}-y^{2}}\).

(c) \(P,Q,R,S\) lie on the circle centre \(O\), with \(\angle POQ=160^\circ\), \(\angle QSR=45^\circ\) and \(\angle PQS=40^\circ\).

(i) \(\angle QPS\): The chord \(PQ\) subtends the central angle \(\angle POQ=160^\circ\). The angle it subtends at the circumference (at \(S\)) is half of this:

\[\angle PSQ=\tfrac{1}{2}\times 160^\circ=80^\circ.\]

In triangle \(PQS\), the three angles sum to \(180^\circ\):

\[\angle QPS=180^\circ-\angle PQS-\angle PSQ=180^\circ-40^\circ-80^\circ=\mathbf{60^\circ}.\]

(ii) \(\angle RQS\): Work out the arcs (central angles) from the given inscribed angles.

• Arc \(PQ\) corresponds to \(\angle POQ=160^\circ\).
• \(\angle PQS=40^\circ\) stands on chord \(PS\), so arc \(PS=2\times40^\circ=80^\circ\).
• \(\angle QSR=45^\circ\) stands on chord \(QR\), so arc \(QR=2\times45^\circ=90^\circ\).

The four arcs around the circle sum to \(360^\circ\):

\[\text{arc }PQ+\text{arc }QR+\text{arc }RS+\text{arc }SP=360^\circ\]\[160^\circ+90^\circ+\text{arc }RS+80^\circ=360^\circ\]\[\text{arc }RS=30^\circ.\]

\(\angle RQS\) stands at \(Q\) on chord \(RS\), so it equals half of arc \(RS\):

\[\angle RQS=\tfrac{1}{2}\times30^\circ=\mathbf{15^\circ}.\]