A farmer owns a triangular field ABC. The length of side [AB] is 85 m and side [AC] is 110 m.The angle between these two sides is 55°.

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

Area$=\frac{1}{2}\times 110\times 85\times \sin 55°$ (M1)(A1)

$=3830\left(3829.53\dots \right){\text{m}}^2$ A1

Note: units must be given for the final A1 to be awarded.

[3 marks]

${\text{BC}}^2={110}^2+{85}^2-2\times 110\times 85\times \cos 55°$ (M1)A1

$\text{BC}=92.7(92.7314\dots )\left(\text{m}\right)$ A1

METHOD 1

Because the height and area of each triangle are equal they must have the samelength base R1

$\text{D}$ must be placed half-way along $\text{BC}$ A1

$\text{BD}=\frac{92.731\dots }{2}\approx 46.4\left(\text{m}\right)$ A1

Note: the final two marks are dependent on the R1 being awarded.

METHOD 2

Let$\text{C}\hat{\text{B}}\text{A}=\theta °$

$\frac{\sin \theta }{110}=\frac{{\displaystyle \sin 55°}}{{\displaystyle 92.731\dots }}$ M1

$\Rightarrow \theta =76.3°\left(76.3354\dots \right)$

Use of area formula

$\frac{1}{2}\times 85\times \text{BD}\times \sin \left(76.33\dots °\right)=\frac{3829.53\dots }{2}$ A1

$\text{BD}=46.4(46.365\dots )\left(\text{m}\right)$ A1

[6 marks]