Intermediate November 2020 Unit 2 Q15
15. In the diagram below, AB, BC and CD are three sides of a regular polygon.
The polygon has 15 sides.
The length of each side is 8 cm.
The exterior angle of the polygon is \(x^\circ\).
BRC is a right-angled triangle.
Diagram not drawn to scale
Calculate the length of BR. [5]
| Scheme | Marks | Comments |
|---|---|---|
| ( x =) \(\dfrac{360}{15}\) or 180 – \(\dfrac{(15 - 2) \times 180}{15}\) or equivalent | M1 | May be seen in parts. |
| = 24(°) | A1 | |
| (BR =) 8 × cos 24 or 8 × sin (90 – 24) | M2 | FT ‘their stated value for x’ ( x < 90°) M1 for \(\dfrac{\text{BR}}{8}\) = cos 24 or \(\dfrac{\text{BR}}{8}\) = sin (90 – 24) Accept equivalent of using sin rule (as sin 90 = 1). |
| = 7·3(0…)(cm) or 7·31(cm) | A1 |
Alternative method to find BR
| Scheme | Marks | Comments |
|---|---|---|
| A correct and complete method (using two trigonometric relationships and possibly Pythagoras’s theorem) | M2 | |
| BR = 7·3(0…)(cm) or 7·31(cm) | A1 |
In the original the alternative method is printed in the Comments column.