Higher June 2021 Paper 2 Q7
7 Here is a right-angled triangle.
Diagram NOT accurately drawn
Work out the value of \(x\).
Give your answer correct to one decimal place.
(3)
| Scheme | Marks |
|---|---|
\(\sin 42 = \dfrac{6.5}{x}\) or \(\dfrac{x}{\sin 90} = \dfrac{6.5}{\sin 42}\) or \(\cos 48 = \dfrac{6.5}{x}\) [where \(48 = 180 - 90 - 42\)] | M1 |
\([x =]\ \dfrac{6.5}{\sin 42}\) or \(\dfrac{6.5 \sin 90}{\sin 42}\) or \([x =]\ \dfrac{6.5}{\cos 48}\) [where \(48 = 180 - 90 - 42\)] | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 9.7 | A1 |
| (3) | |
| (3 marks) |
Notes
M1: or use of tan to find the horizontal side and then a correct first step in Pythagoras’ theorem ie [base =] \(\dfrac{6.5}{\tan 42}\ (= 7.21\ldots)\) and \([x^2 =]\ 6.5^2 + \text{``}{7.21\ldots}\text{''}^2\)
M1: or complete method using Pythagoras \([x =]\ \sqrt{6.5^2 + \text{``}{7.21\ldots}\text{''}^2}\)
(If students give this statement with nothing before it they gain M2)
A1: accept 9.7 – 9.72