Find the length of the path BD.

Show that angle DBC is 48.7°, correct to three significant figures.

Find the area of the park.

Find the length of the path CE.

Calculate the total length of the track.

$({\text{B}}{{\text{D}}^2} = ){\text{ }}{95^2} + {120^2} - 2 \times 95 \times 120 \times \cos 70^\circ$     (M1)(A1)

Note:     Award (M1) for substituted cosine rule, (A1) for correct substitution.

$({\text{BD}} = ){\text{ }}125{\text{ (m) }}\left( {125.007 \ldots {\text{ (m)}}} \right)$     (A1)(G2)

[3 marks]

$\frac{{\sin {\text{DBC}}}}{{100}} = \frac{{\sin 110^\circ }}{{125.007 \ldots }}$     (M1)(A1)(ft)

Note:     Award (M1) for substituted sine rule, (A1)(ft) for correct substitution.

Follow through from their answer to part (a).

$({\text{DBC}} = ){\text{ }}48.7384 \ldots ^\circ$     (A1)(ft)

$({\text{DBC}} = ){\text{ }}48.7^\circ$     (AG)

Notes:     Award the final (A1)(ft) only if both their unrounded answer and 48.7° is seen. Follow through from their answer to part (a), only if their unrounded answer rounds to 48.7°.

[3 marks]

$\frac{1}{2} \times 125.007 \ldots  \times 100 \times \sin 21.3^\circ  + \frac{1}{2} \times 95 \times 120 \times \sin 70^\circ$     (A1)(M1)(M1)

Note:     Award (A1) for 21.3° (21.2615…) seen, (M1) for substitution into (at least) one area of triangle formula in the form $\frac{1}{2}ab\sin c$, (M1) for their correct substitutions and adding the two areas.

$7630{\text{ }}{{\text{m}}^2}{\text{ }}(7626.70 \ldots {{\text{m}}^2})$     (A1)(ft)(G3)

Notes:     Follow through from their answers to part (a). Accept $7620{\text{ }}{{\text{m}}^2}{\text{ }}(7622.79 \ldots {{\text{m}}^2})$ from use of 48.7384…

[4 marks]

$({\text{CE}} = ){\text{ }}100 \times \sin 21.3^\circ$     (M1)

$({\text{CE}} = ){\text{ }}36.3{\text{ (m) }}\left( {36.3251 \ldots {\text{ (m)}}} \right)$     (A1)(ft)(G2)

Note:     Follow through from their angle 21.3° in part (c). Award (M0)(A0) for halving 110° and/or assuming E is the midpoint of BD in any method seen.

OR

${\text{area of BCD}} = \frac{1}{2}{\text{BD}} \times {\text{CE}}$     (M1)

$({\text{CE}} = ){\text{ }}36.3{\text{ (m) }}\left( {36.3251 \ldots {\text{ (m)}}} \right)$     (A1)(ft)(G2)

Note:     Follow through from parts (a) and (c). Award (M0)(A0) for halving 110° and/or assuming E is the midpoint of BD in any method seen.

[2 marks]

$\sqrt {{{100}^2} - 36.3251{ \ldots ^2}}  + 100 + 36.3251 \ldots$     (M1)(M1)

Note:     Award (M1) for correct use of Pythagoras to find DE (or correct trigonometric equation, $100 \times \cos 21.3$, to find DE), (M1) for the sum of 100, their DE and their CE.

$229{\text{ (m) }}\left( {229.494 \ldots {\text{ (m)}}} \right)$     (A1)(ft)(G2)

Note:     Follow through from part (d). Use of 3 sf values gives an answer of $230{\text{ (m) }}\left( {229.5{\text{ (m)}}} \right)$.

[3 marks]