(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 the circumference. Prove that < AOB = 2 < APB.
(ii)
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)
(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°)