Higher June 2023 Paper 2 Q13
13 \(ABC\) is a triangle.
Calculate the size of angle \(BAC\).
Give your answer correct to 1 decimal place. (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 58.5 | P1 | for start of process to find angle \(BCA\), eg \(\dfrac{18}{\sin 70} = \dfrac{15}{\sin BCA}\) or \(\dfrac{\sin 70}{18} = \dfrac{\sin BCA}{15}\) |
| P1 | for rearrangement, eg \((\sin BCA =)\ \dfrac{15 \sin 70}{18}\ (= 0.783\ldots)\) oe or \(BCA = 51.5\ldots\) | |
| P1 | for complete process to find angle \(BAC\), eg \(180 - 70 - \sin^{-1}\left(\dfrac{15 \sin 70}{18}\right)\) | |
| A1 | for answer in the range 58.4 to 58.5 |
Additional guidance
Angle \(BCA\) must be correctly identified to gain marks
\(\sin 70 = 0.939\ldots\)
\(\sin 70 \div 18 = 0.052\ldots\) \(18 \div \sin 70 = 19.1\ldots\)
If an answer is given in the range in working and then rounded incorrectly award full marks.