Find the size of angle $ABC$.

Find the area of the triangular field.

${\text{M}}$ is the midpoint of ${\text{AC}}$.

Find the length of ${\text{BM}}$.

A vertical mobile phone mast, ${\text{TB}}$, is built next to the field with its base at ${\text{B}}$. The angle of elevation of ${\text{T}}$ from ${\text{M}}$ is $63.4^\circ$. ${\text{N}}$ is the midpoint of the mast.

Calculate the angle of elevation of ${\text{N}}$ from ${\text{M}}$.

$\frac{{70}}{{\sin 78}} = \frac{{50}}{{\sin {\rm{A\hat BC}}}}$     (M1)(A1)

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

${\rm{A\hat BC}} = 44.3^\circ$ ($44.3209...$)     (A1)(G3)

Note: If radians are used the answer is $0.375918...$, award at most (M1)(A1)(A0).

[3 marks]

${\text{area }}\Delta {\text{ABC}} = \frac{1}{2} \times 70 \times 50 \times \sin (57.6790 \ldots )$     (A1)(M1)(A1)(ft)
 

Notes: Award (A1)(ft) for their $57.6790…$ seen, (M1) for substituted area formula, (A1)(ft) for correct substitution.

     Follow through from part (a).

$= 1480{\text{ }}{{\text{m}}^2}$  $(1478.86 \ldots )$     (A1)(ft)(G3)

Notes: The answer is $1480{\text{ }}{{\text{m}}^2}$, units are required. $1479.20…$ if 3 sf used.

     If radians are used the answer is $1554.11 \ldots {{\text{m}}^2}$, award (A1)(ft)(M1)(A1)(ft)(A1)(ft)(G3).

[4 marks]

${\text{B}}{{\text{M}}^2} = {70^2} + {25^2} - 2 \times 70 \times 25 \times \cos (57.6790 \ldots )$     (M1)(A1)(ft)

Notes: Award (M1) for substituted cosine rule, (A1)(ft) for correct substitution. Follow through from their angle in part (b).

${\text{BM}} = 60.4{\text{ (m)}}$   $(60.4457 \ldots )$     (A1)(ft)(G2)

Notes: If the 3 sf answer is used the answer is $60.5{\text{ }}({\text{m}})$.

     If radians are used the answer is $62.5757… {\text{ }} ({\text{m}})$, award (M1)(A1)(ft)(A1)(ft)(G2).

[3 marks]

$\tan 63.4^\circ  = \frac{{{\text{TB}}}}{{60.4457 \ldots }}$     (M1)

Note: Award (M1) for their correctly substituted trig equation.

${\text{TB}} = 120.707 \ldots$     (A1)(ft)

Notes: Follow through from part (c). If 3 sf answers are used throughout, ${\text{TB}} = 120.815 \ldots$

If ${\text{TB}} = 120.707 \ldots$ is seen without working, award (A2).

$\tan {\rm{N\hat MB = }}\frac{{\left( {\frac{{120.707 \ldots }}{2}} \right)}}{{60.4457 \ldots }}$     (A1)(ft)(M1)

Notes: Award (A1)(ft) for their ${\text{TB}}$ divided by $2$ seen, (M1) for their correctly substituted trig equation.

     Follow through from part (c) and within part (d).

${\rm{N\hat MB = 45.0^\circ }}$   ${\text{(44.9563}} \ldots {\text{)}}$     (A1)(ft)(G3)

Notes:     If 3 sf are used throughout, answer is $45^\circ$.

     If radians are used the answer is $0.308958…$, and if full working is shown, award at most (M1)(A1)(ft)(A1)(ft)(M1)(A0).

     If no working is shown for radians answer, award (G2).

OR

$\tan {\rm{N\hat MB}} = \frac{{{\text{NB}}}}{{{\text{BM}}}}$     (M1)

$\tan 63.4^\circ  = \frac{{2 \times {\text{NB}}}}{{{\text{BM}}}}$     (A1)(M1)

Note: Award (A1) for $2 \times {\text{NB}}$ seen.

$\tan {\rm{N\hat MB}} = \frac{1}{2}\tan 63.4^\circ$     (M1)

${\rm{N\hat MB}} = 45.0^\circ$   $(44.9563 \ldots )$     (A1)(G3)

Notes: If radians are used the answer is $0.308958…$, and if full working is shown, award at most (M1)(A1)(M1)(M1)(A0). If no working is shown for radians answer, award (G2).

[5 marks]