(a) In the diagram, MNPQ is a circle with centre O, |MN| = |NP| and < OMN = 50°. Find: (I) < MNP (ii) < POQ (b) Find the equation of the line which has the ...

Answer Details

(a) Circle \(MNPQ\) with centre \(O\), \(|MN|=|NP|\) and \(\angle OMN=50^{\circ}\).

From the diagram \(M\) and \(Q\) are the ends of a diameter through \(O\), and the points lie in the order \(M, N, P, Q\) round the circle.

(i) Find \(\angle MNP\).

\(OM\) and \(ON\) are radii, so triangle \(OMN\) is isosceles with \(OM=ON\). Hence the base angles are equal:

\[\angle ONM=\angle OMN=50^{\circ}\]

Angle at the centre of triangle \(OMN\):

\[\angle MON=180^{\circ}-50^{\circ}-50^{\circ}=80^{\circ}\]

Equal chords subtend equal angles at the centre, and \(|MN|=|NP|\), so:

\[\angle NOP=\angle MON=80^{\circ}\]

The central angle standing on the minor arc \(MNP\) is therefore

\[\angle MON+\angle NOP=80^{\circ}+80^{\circ}=160^{\circ}.\]

The reflex central angle on the major arc \(MP\) (the arc not containing \(N\)) is

\[360^{\circ}-160^{\circ}=200^{\circ}.\]

The inscribed angle \(\angle MNP\) stands on this major arc \(MP\), so it is half of it:

\[\angle MNP=\frac{1}{2}\times 200^{\circ}=100^{\circ}\]

(ii) Find \(\angle POQ\).

\(M\), \(O\), \(Q\) are collinear (diameter), so the central angles along the arc \(M\to N\to P\to Q\) add up to a straight angle:

\[\angle MON+\angle NOP+\angle POQ=180^{\circ}\]\[80^{\circ}+80^{\circ}+\angle POQ=180^{\circ}\]\[\angle POQ=20^{\circ}\]

(b) Line with the same gradient as \(8y+4x=24\) through \((-8,\,12)\).

Make \(y\) the subject to read off the gradient:

\[8y=24-4x\]\[y=3-\tfrac{1}{2}x\]

The gradient is \(m=-\dfrac{1}{2}\). Using \(y-y_1=m(x-x_1)\) with \((x_1,y_1)=(-8,12)\):

\[y-12=-\tfrac{1}{2}(x+8)\]\[y-12=-\tfrac{1}{2}x-4\]\[y=-\tfrac{1}{2}x+8\]

The required line is \(y=-\dfrac{1}{2}x+8\), i.e. \(x+2y=16\).