Write down the size of angle ABC.

Calculate the length of AC.

Calculate the area of the field ABC.

N is the point on AB such that CN is perpendicular to AB. M is the midpoint of CN.

Calculate the length of NM.

A goat is attached to one end of a rope of length 7 m. The other end of the rope is attached to the point M.

Decide whether the goat can reach point P, the midpoint of CB. Justify your answer.

\({\text{Angle ABC}} = {50^ \circ }\)     (A1)

[1 mark]

\(\frac{{{\text{AC}}}}{{\sin {{50}^ \circ }}} = \frac{{25}}{{\sin {{55}^ \circ }}}\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for correct substitution. Follow through from their angle ABC.

\({\text{AC}} = 23.4{\text{ m}}\)     (A1)(ft)(G2)

[3 marks]

\({\text{Area of }}\Delta {\text{ ABC}} = \frac{1}{2} \times 23.379 \ldots  \times 25 \times \sin {75^ \circ }\)     (M1)(A1)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for correct substitution. Follow through from their AC.

OR

\({\text{Area of triangle ABC}} = \frac{{29.479 \ldots  \times 19.151 \ldots }}{2}\)     (A1)(ft)(M1)

Note: (A1)(ft) for correct values of AB (29.479…) and CN (19.151…). Follow through from their (a) and /or (b). Award (M1) for substitution of their values of AB and CN into the correct formula.

\({\text{Area of }}\Delta {\text{ ABC}} = 282{\text{ }}{{\text{m}}^2}\)     (A1)(ft)(G2)

Note: Accept \(283{\text{ }}{{\text{m}}^2}\) if \(23.4\) is used.

[3 marks]

\({\text{NM}} = \frac{{25 \times \sin {{50}^ \circ }}}{2}\)     (M1)(M1)

Note: Award (M1) for \({25 \times \sin {{50}^ \circ }}\) or equivalent for the length of CN. (M1) for dividing their CN by \(2\).

\({\text{NM}} = 9.58{\text{ m}}\)     (A1)(ft)(G2)

Note: Follow through from their angle ABC.

Notes: Premature rounding of CN leads to the answers \(9.60\) or \(9.6\). Award at most (M1)(M1)(A0) if working seen. Do not penalize with (AP). CN may be found in (c).

Note: The working for this part of the question may be in part (b).

[3 marks]

\({\text{Angle NCB}} = {40^ \circ }\) seen     (A1)(ft)

Note: Follow through from their (a).

From triangle MCP:

\({\text{M}}{{\text{P}}^2} = {(9.5756 \ldots )^2} + {12.5^2} - 2 \times 9.5756 \ldots  \times 12.5 \times \cos ({40^ \circ })\)     (M1)(A1)(ft)

\({\text{MP}} = 8.034 \ldots {\text{ m}}\)     (A1)(ft)(G3)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their d). Award (G3) for correct value of MP seen without working.

OR

From right triangle MCP

\({\text{CP}} = 12.5{\text{ m}}\) seen     (A1)

\({\text{M}}{{\text{P}}^2} = {(12.5)^2} - {(9.575 \ldots )^2}\)     (M1)(A1)(ft)

\({\text{MP}} = 8.034 \ldots {\text{ m}}\)     (A1)(G3)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their (d). Award (G3) for correct value of MP seen without working.

OR

From right triangle MCP

\({\text{Angle MCP}} = {40^ \circ }\) seen     (A1)(ft)

\(\frac{{{\text{MP}}}}{{12.5}} = \sin ({40^ \circ })\) or equivalent     (M1)(A1)(ft)

\({\text{MP}} = 8.034 \ldots {\text{ m}}\)     (A1)(G3)(ft)

Notes: Award (M1) for substitution into the correct formula, (A1)(ft) for their correct substitution. Follow through from their (a). Award (G3) for correct value of MP seen without working.

The goat cannot reach point P as \({\text{MP}} > 7{\text{ m}}\) .     (A1)(ft)

Note: Award (A1)(ft) only if their value of MP is compared to \(7{\text{ m}}\), and conclusion is stated.

[5 marks]

Many candidates assumed incorrectly that the triangle ABC is isosceles or/and that CN is an angle bisector, and those assumptions led them to use incorrect methods. Wherever those assumptions were made first, all or most of the marks were lost in that specific part of Question 4. There were provisions in the mark scheme to follow through in subsequent parts. Most candidates at least attempted parts a), b) and c). Some candidates incorrectly used an area formula for a right triangle in c) and lost all marks. Many candidates lost a mark for premature rounding in d). Part e) proved to be especially difficult for the candidates. Here many candidates offered guesses instead of sound mathematical reasoning.

Many candidates assumed incorrectly that the triangle ABC is isosceles or/and that CN is an angle bisector, and those assumptions led them to use incorrect methods. Wherever those assumptions were made first, all or most of the marks were lost in that specific part of Question 4. There were provisions in the mark scheme to follow through in subsequent parts. Most candidates at least attempted parts a), b) and c). Some candidates incorrectly used an area formula for a right triangle in c) and lost all marks. Many candidates lost a mark for premature rounding in d). Part e) proved to be especially difficult for the candidates. Here many candidates offered guesses instead of sound mathematical reasoning.

Many candidates assumed incorrectly that the triangle ABC is isosceles or/and that CN is an angle bisector, and those assumptions led them to use incorrect methods. Wherever those assumptions were made first, all or most of the marks were lost in that specific part of Question 4. There were provisions in the mark scheme to follow through in subsequent parts. Most candidates at least attempted parts a), b) and c). Some candidates incorrectly used an area formula for a right triangle in c) and lost all marks. Many candidates lost a mark for premature rounding in d). Part e) proved to be especially difficult for the candidates. Here many candidates offered guesses instead of sound mathematical reasoning.

Many candidates assumed incorrectly that the triangle ABC is isosceles or/and that CN is an angle bisector, and those assumptions led them to use incorrect methods. Wherever those assumptions were made first, all or most of the marks were lost in that specific part of Question 4. There were provisions in the mark scheme to follow through in subsequent parts. Most candidates at least attempted parts a), b) and c). Some candidates incorrectly used an area formula for a right triangle in c) and lost all marks. Many candidates lost a mark for premature rounding in d). Part e) proved to be especially difficult for the candidates. Here many candidates offered guesses instead of sound mathematical reasoning.

Many candidates assumed incorrectly that the triangle ABC is isosceles or/and that CN is an angle bisector, and those assumptions led them to use incorrect methods. Wherever those assumptions were made first, all or most of the marks were lost in that specific part of Question 4. There were provisions in the mark scheme to follow through in subsequent parts. Most candidates at least attempted parts a), b) and c). Some candidates incorrectly used an area formula for a right triangle in c) and lost all marks. Many candidates lost a mark for premature rounding in d). Part e) proved to be especially difficult for the candidates. Here many candidates offered guesses instead of sound mathematical reasoning.