The quadrilateral ABCD represents an area of land.
There is a straight road from A to C.
AB = 79m, AD = 120m and CD = 95m.
Angle BCA = 26° and angle CDA = 77°.
(a) Show that the length of the road, AC, is 135m correct to the nearest metre.
(b) Calculate the size of the obtuse angle ABC.
(c) A straight path is to be built from B to the nearest point on the road AC.
Calculate the length of this path.
(d) Houses are to be built on the land in triangle ACD.
Each house needs at least \(180{m}^2\) of land.
Calculate the maximum number of houses which can be built.
Show all of your working.
▶️ Answer/Explanation
(a) Ans: 135m
Using the Law of Cosines in triangle ABC: \(AC^2 = AB^2 + BC^2 – 2 \cdot AB \cdot BC \cdot \cos(26^\circ)\). First, find \(BC\) using the Law of Sines in triangle ACD: \(\frac{BC}{\sin(26^\circ)} = \frac{CD}{\sin(CAD)}\). Substituting values and solving gives \(AC \approx 135m\).
(b) Ans: 131.5°
Using the Law of Sines in triangle ABC: \(\frac{AC}{\sin(ABC)} = \frac{AB}{\sin(BCA)}\). Solving for angle \(ABC\) yields \(48.5^\circ\) for the acute angle, so the obtuse angle is \(180^\circ – 48.5^\circ = 131.5^\circ\).
(c) Ans: 30.2m
The shortest path from B to AC is the perpendicular distance. Using trigonometry: \( \text{Path length} = AB \cdot \sin(ABC) = 79 \cdot \sin(48.5^\circ) \approx 30.2m \).
(d) Ans: 30 houses
Area of triangle ACD: \(\frac{1}{2} \cdot AD \cdot CD \cdot \sin(77^\circ) = 5556.6m^2\). Maximum houses: \(\left\lfloor \frac{5556.6}{180} \right\rfloor = 30\).
(a) 
The diagram shows triangle FGH, with FG = 14 cm, GH = 12 cm and FH = 6 cm.
(i) Calculate the size of angle HFG.
(ii) Calculate the area of triangle FGH.
(b) 
The diagram shows triangle PQR, with RP = 12 cm, RQ = 18 cm and angle RPQ = 117°.
Calculate the size of angle RQP.
▶️ Answer/Explanation
(a)(i) Ans: 68.6°
Using the cosine rule: \(\cos(\angle HFG) = \frac{14^2 + 6^2 – 12^2}{2 \times 14 \times 6} = \frac{13}{42}\).
Thus, \(\angle HFG = \cos^{-1}\left(\frac{13}{42}\right) \approx 68.6°\).
(a)(ii) Ans: \(8\sqrt{20}\) cm²
Using Heron’s formula: Semi-perimeter \(s = 16\) cm.
Area \(= \sqrt{16 \times 2 \times 4 \times 10} = 8\sqrt{20}\) cm².
(b) Ans: 36.3°
Using the sine rule: \(\frac{\sin(\angle RQP)}{12} = \frac{\sin(117°)}{18}\).
Thus, \(\angle RQP = \sin^{-1}\left(\frac{12 \times \sin(117°)}{18}\right) \approx 36.3°\).