Calculate the length of AC.

Calculate the length of AG.

Calculate the angle that AG makes with the floor.

\({\text{A}}{{\text{C}}^2} = {7.2^2} + {9.6^2}\)     (M1) 

Note: Award (M1) for correct substitution in Pythagoras Theorem.

\({\text{AC}} = 12{\text{ m}}\)     (A1)     (C2)

[2 marks]

\({\text{A}}{{\text{G}}^2} = {12^2} + {3.5^2}\)     (M1) 

Note: Award (M1) for correct substitution in Pythagoras Theorem.

\({\text{AG}} = 12.5{\text{ m}}\)     (A1)(ft)     (C2)

Note: Follow through from their answer to part (a).

[2 marks]

\(\tan \theta  = \frac{{3.5}}{{12}}\) or \(\sin \theta  = \frac{{3.5}}{{12.5}}\) or \(\cos \theta  = \frac{{12}}{{12.5}}\)     (M1)

Note: Award (M1) for correct substitutions in trig ratio.

\(\theta  = {16.3^ \circ }\)     (A1)(ft)     (C2)

Notes: Follow through from parts (a) and/or part (b) where appropriate. Award (M1)(A0) for use of radians (0.284).

[2 marks]

Question 7 was surprisingly difficult for many candidates, especially part b. Many candidates did not recognize that ACG was a right angled triangle and tried to use the law of cosines to find angle A. Although correct substitution and manipulation provided the correct answer, many candidates attempting this method made arithmetical errors.

Question 7 was surprisingly difficult for many candidates, especially part b. Many candidates did not recognize that ACG was a right angled triangle and tried to use the law of cosines to find angle A. Although correct substitution and manipulation provided the correct answer, many candidates attempting this method made arithmetical errors.

Question 7 was surprisingly difficult for many candidates, especially part b. Many candidates did not recognize that ACG was a right angled triangle and tried to use the law of cosines to find angle A. Although correct substitution and manipulation provided the correct answer, many candidates attempting this method made arithmetical errors.