(a) Make d the subject of the formula \(S = \frac{n}{2}[2a + (n - 1)
(a) Make d the subject of the formula \(S = \frac{n}{2}[2a + (n - 1) d]\).
(b) (i).jpg)
In the diagram, O is the centre of the circle, A, B and P are points on the circumference. Prove that < AOB = 2 < APB.
(ii).jpg)
Find the angles x, v and z in the diagram.
Explanation
(a) \(S = \frac{n}{2} [2a + (n - 1)d]\)
\(2S = n[2a + (n - 1) d]\)
\(\frac{2S}{n} = 2a + (n - 1)d\)
\(\frac{2S}{n} - 2a = (n - 1) d\)
\(d = \frac{\frac{2S}{n} - 2a}{(n - 1)}\)
= \(\frac{2S - 2an}{n(n - 1)}\)
(b).jpg)
(i) Given: circle ABP
To prove < AOB = 2 < APB
Construction : Join PO produced to Q.
Proof : /OA/ = /OB/ (radii)
\(\therefore x_{1} = x_{2} \) (base angles of isosceles triangle)
\(\therefore < AOQ = x_{1} + x_{2}\) (exterior angle of triangle AOP)
\(\therefore < AOQ = 2x_{2} (x_{1} = x_{2})\)
Similarly, < AOB = < BOQ - < AOQ.
= \(2y_{2} - 2x_{2} = 2(y_{2} - x_{2}) = 2 \times < APB\)
\(\therefore < AOB = 2 < APB.\)
(ii) \(z = 2x \) (angle subtended at the centre)
\(z = 2(126°) = 252°\)
\(\therefore x = 360° - 252° = 108°\)
\(\therefore y = \frac{1}{2} \times 108° = 54°\)
\(x = 108° ; y = 54° ; z = 252°\)