Higher June 2019 Paper 6 Q5
5 ABC is a right-angled triangle.
AB = 20 cm and BC = 37 cm.
Not to scale
Calculate angle BAC. [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Answer which rounds to 61.6 nfww | 3 | M2 for \(\tan^{-1}\left(\dfrac{37}{20}\right)\) oe or M1 for tan[x =] \(\dfrac{37}{20}\) oe If M0 scored then SC1 for answers 28.4, 28 or angles that round to 28.4 if correct working seen. | Condone answer of 62 only if correct working seen Answers of 68.5 or 68.4(5..) [grads] or 1.08 or 1.07(5..) [rads] imply M2 Alternative method After correct method for Pythagoras soi by 42.0 to 42.1 M2 for \(\sin^{-1}\left(\dfrac{37}{\textit{their}\sqrt{20^2 + 37^2}}\right)\) or \(\cos^{-1}\left(\dfrac{20}{\textit{their}\sqrt{20^2 + 37^2}}\right)\) or M1 for sin[x =] \(\dfrac{37}{\textit{their}\sqrt{20^2 + 37^2}}\) or cos[x =] \(\dfrac{20}{\textit{their}\sqrt{20^2 + 37^2}}\) or M0 for just Pythagoras reaching AC = 42.0 to 42.1 Do not condone answer of 62 following an interim answer seen that does not round to 61.6 0 for scale drawing |