Calculate the length of AB .

M is the midpoint of EF .

Calculate the length of BM .

M is the midpoint of EF .

Find the size of the angle between BM and the face ADEF .

9.5² = 8² + AB²     (M1)

Note: Award (M1) for correct substitution into Pythagoras’ theorem.

AB = 5.12 (cm) (5.12347…)     (A1)     (C2)

[2 marks]

\({\text{BM = }}\sqrt {{{9.5}^2} + {{\left( {\frac{{5.12347...}}{2}} \right)}^2}} \)     (M1)

Note: Award (M1) for correct substitution into Pythagoras’ theorem.

= 9.84 (cm) (9.83933…)     (A1)(ft)     (C2)

Notes: Accept alternative methods. Follow through from their answer to part

[2 marks]

sin \({\text{A}}\hat {\rm M}{\text{B = }}\frac{{5.12347...}}{{9.83933...}}\)     (M1)

Note: Award (M1) for a correctly substituted trigonometrical equation using \({\text{A}}\hat {\rm M}{\text{B}}\) .

= 31.4 (31.3801...)     (A1)(ft)     (C2)

Notes: If radians used, the answer will be 0.5476… award (M1)(A0)(ft). Degree symbol ° not required. Follow through from their answers to part (a) and to part (b).

[2 marks]

This seemed to be a good discriminatory question enabling the majority of candidates to at least score well on part (a). Challenges arose for candidates who were then required to see the problem in three dimensions for the remainder of the question. Indeed, a significant number of candidates correctly identified the required lengths for part (b) and, provided they used Pythagoras correctly, were able to pick up the marks in this part of the question. However, in part (c), invariably the wrong triangle was chosen with triangles BFM and BAF proving to be the most popular, but incorrect triangles, chosen.

This seemed to be a good discriminatory question enabling the majority of candidates to at least score well on part (a). Challenges arose for candidates who were then required to see the problem in three dimensions for the remainder of the question. Indeed, a significant number of candidates correctly identified the required lengths for part (b) and, provided they used Pythagoras correctly, were able to pick up the marks in this part of the question. However, in part (c), invariably the wrong triangle was chosen with triangles BFM and BAF proving to be the most popular, but incorrect triangles, chosen.

This seemed to be a good discriminatory question enabling the majority of candidates to at least score well on part (a). Challenges arose for candidates who were then required to see the problem in three dimensions for the remainder of the question. Indeed, a significant number of candidates correctly identified the required lengths for part (b) and, provided they used Pythagoras correctly, were able to pick up the marks in this part of the question. However, in part (c), invariably the wrong triangle was chosen with triangles BFM and BAF proving to be the most popular, but incorrect triangles, chosen.