Intermediate November 2020 Unit 2 Q19
19. The diagram below shows a circle with centre at point O.
A, B, C and D are all points on the circumference of the circle.
AB = 7·5 cm and BC = 4·7 cm.
Diagram not drawn to scale
(a)
(i) Give the reason why \(A\hat{B}C\) is 90°. [1]
(ii) Calculate the size of angle \(x\). [3]
(b) Write down the size of angle \(y\).
State the circle theorem you have used to find your answer. [2]
State the circle theorem you have used to find your answer. [2]
\(y\) = ……………………
Circle theorem used: ……………………
| Scheme | Marks | Comments |
|---|---|---|
| (i) Correct reason given. e.g. ‘An angle at the circumference subtended by a diameter is a right angle’. ‘ line AC is a diameter’ | E1 | Accept any correct unambiguous wording. The key word is ‘diameter’. Allow eg ‘angle in a semicircle is 90°’, ‘line AC goes through the centre’. ‘opposite a diameter’ Do not accept ‘because it’s a right angle’. |
| (ii) tan x = \(\dfrac{7{\cdot}5}{4{\cdot}7}\) | M1 | |
| x = tan−1 (7·5 / 4·7) or tan−1 1.6 or tan−1 1.59(…) | m1 | Implies M1. |
| = 57·9(…)(°) or 57·8(…)(°) or 58(°) | A1 | C.A.O. |
Alternative method to find x
| Scheme | Marks | Comments |
|---|---|---|
| A correct and complete method (using Pythagoras’s theorem and a trigonometric relationship). | M2 | |
| x = 57·9(…)(°) or 57·8(…)(°) or 58(°) CAO | A1 |
In the original the alternative method is printed in the Comments column.
| Scheme | Marks | Comments |
|---|---|---|
| (y =) 58(°) | B1 | Strict FT of ‘their x’. |
| Correct circle theorem given. e.g. ‘angles (at the circumference) subtended by the same chord (or arc) are equal’, ‘angles in the same segment (are equal)’. | E1 | Accept any correct unambiguous wording. Allow eg ‘angles on the same chord (are equal)’ Do not accept e.g. ‘they are equal’ on its own. |