Find the size of BĈA.

Calculate the length of AD.

The farmer wants to build a fence around ABD.

Calculate the total length of the fence.

The farmer wants to build a fence around ABD.

The farmer pays 802.50 USD for the fence. Find the cost per metre.

Calculate the area of the triangle ABD.

A layer of earth 3 cm thick is removed from ABD. Find the volume removed in cubic metres.

\(\frac{{\sin {\text{BCA}}}}{{35}} = \frac{{\sin 105^\circ }}{{80}}\)     (M1)(A1)

Note: Award (M1) for correct substituted formula, (A1) for correct substitutions.

\({\text{B}}{\operatorname{\hat C}}{\text{A}} = 25.0^{\circ}\)     (A1)(G2)

[3 marks]

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

Length BD = 40 m     (A1)

Angle ABC = 180° − 105° − 25° = 50°     (A1)(ft)

Note: (ft) from their answer to (a).

AD² = 35² + 40² − (2 × 35 × 40 × cos 50°)     (M1)(A1)(ft)

Note: Award (M1) for correct substituted formula, (A1)(ft) for correct substitutions.

(UP)     AD = 32.0 m     (A1)(ft)(G3)

Notes: If 80 is used for BD award at most (A0)(A1)(ft)(M1)(A1)(ft)(A1)(ft) for an answer of 63.4 m.

If the angle ABC is incorrectly calculated in this part award at most (A1)(A0)(M1)(A1)(ft)(A1)(ft).

If angle BCA is used award at most (A1)(A0)(M1)(A0)(A0).

[5 marks]

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

length of fence = 35 + 40 + 32     (M1)

(UP)     = 107 m     (A1)(ft)(G2)

Note: (M1) for adding 35 + 40 + their (b).

[2 marks]

cost per metre \( = \frac{802.50}{107}\)     (M1)

Note: Award (M1) for dividing 802.50 by their (c).

cost per metre = 7.50 USD (7.5 USD) (USD not required)     (A1)(ft)(G2)

[2 marks]

Note: Unit penalty (UP) applies in parts (b)(c) and (e)

Area of ABD \( = \frac{1}{2} \times 35 \times 40 \times \sin 50^\circ \)     (M1)

= 536.2311102     (A1)(ft)

(UP)     = 536 m²     (A1)(ft)(G2)

Note: Award (M1) for correct substituted formula, (A1)(ft) for correct substitution, (ft) from their value of BD and their angle ABC in (b).

[3 marks]

Volume = 0.03 × 536     (A1)(M1)

= 16.08

= 16.1     (A1)(ft)(G2)

Note: Award (A1) for 0.03, (M1) for correct formula. (ft) from their (e).

If 3 is used award at most (A0)(M1)(A0).

[3 marks]

This was a simple application of non-right angled trigonometry and most candidates answered it well. Some candidates lost marks in both parts due to the incorrect setting of the calculators. Those that did not score well overall primarily used Pythagoras.

This was a simple application of non-right angled trigonometry and most candidates answered it well. Some candidates lost marks in both parts due to the incorrect setting of the calculators. Those that did not score well overall primarily used Pythagoras.

Most candidates scored full marks, many by follow through from an incorrect part (b). The main error was using the value for BC and not BD.

Most candidates scored full marks, many by follow through from an incorrect part (b). The main error was using the value for BC and not BD.

Done well; again some candidates used the right-angled formula.

This part was poorly done; many candidates unable to convert 3 cm to 0.03 m. A significant number used the wrong formula, multiplying their answer by 1/3.