Higher November 2018 Paper 3 Q12
12 Here is a pyramid with a square base \(ABCD\).
\(AB = 5\) m
The vertex \(T\) is 12 m vertically above the midpoint of \(AC\).
Calculate the size of angle \(TAC\). (4)
| Answer | Mark | Mark scheme |
|---|---|---|
| 73.6 | P1 | for correct initial use of Pythagoras eg \(5^2 + 5^2\) (=50) or a trigonometric ratio in the form \(\dfrac{5 \div 2}{0.5AC} = \sin 45\) oe |
| P1 | for finding the length of half of the diagonal eg \(\sqrt{\text{``}50\text{''}} \div 2\) ( = 3.5...) or \(0.5AC = \dfrac{5 \div 2}{\sin 45}\) (=3.5...) oe | |
| P1 | for process to use tan eg \(\tan TAC = (12 \div \text{``}3.5..\text{''})\) (=3.3..) or complete alternative method arriving at an equation with the subject as \(\sin TAC\) or \(\cos TAC\) | |
| A1 | for an answer in the range 73.58 to 74.1 |
Additional guidance
P1 (half of the diagonal): do not accept \(\sqrt{20} \div 2\)