(a) Make d the subject of the formula \(S = \frac{n}{2}[2a + (n - 1) d]\). (b) (i) In the diagram, O is the centre of the circle, A, B and P are points on t...

Answer Details

(a) Make d the subject of \(S=\frac{n}{2}[2a+(n-1)d]\).

\[2S=n[2a+(n-1)d]\]\[\frac{2S}{n}=2a+(n-1)d\]\[\frac{2S}{n}-2a=(n-1)d\]\[d=\frac{\dfrac{2S}{n}-2a}{n-1}=\frac{2S-2an}{n(n-1)}=\frac{2(S-an)}{n(n-1)}.\]

(b)(i) Prove that \(\angle AOB=2\angle APB\).

Given: circle with centre O; A, B and P on the circumference.
To prove: \(\angle AOB=2\angle APB\).
Construction: Join P to O and produce it to a point Q.
Proof: \(OA=OP\) (radii), so triangle OAP is isosceles and its base angles are equal, \(x_1=x_2\). The exterior angle of a triangle equals the sum of the two interior opposite angles, so

\[\angle AOQ=x_1+x_2=2x_2.\]

Similarly, \(OB=OP\) (radii), giving \(\angle BOQ=2y_2\) where \(y_2=\angle OPB\). Adding the two central angles:

\[\angle AOB=\angle AOQ+\angle BOQ=2x_2+2y_2=2(x_2+y_2)=2\angle APB.\]\[\therefore\ \angle AOB=2\angle APB.\]

(b)(ii) Find x, y and z. The inscribed angle standing on the same arc is \(126^\circ\). The reflex angle at the centre is twice the angle at the circumference:

\[z=2\times126^\circ=252^\circ.\]

The angle at the centre going the other way completes the full turn:

\[x=360^\circ-252^\circ=108^\circ.\]

The angle at the circumference on this arc is half the central angle:

\[y=\tfrac{1}{2}\times108^\circ=54^\circ.\]\[\therefore\ x=108^\circ,\quad y=54^\circ,\quad z=252^\circ.\]