The diagram shows triangle ABC on horizontal ground.
AC = 15 m, BC = 8 m and AB = 20 m.
BP and CQ are vertical poles of different heights.
BP = 3 m and CQ = 4m.
AQ and PQ are straight wires.
(a) Show that angle ACB = 117.5°, correct to 1 decimal place.
(b) Calculate the area of triangle ABC.
(c) Calculate the length of AQ.
(d) Calculate the angle of elevation of Q from P.
(e) Another straight wire connects A to the midpoint of PQ.
Calculate the angle between this wire and the horizontal ground.
▶️ Answer/Explanation
(a) Using cosine rule: cos(ACB) = (15² + 8² – 20²)/(2×15×8) = -111/240 ≈ -0.4625
ACB = cos⁻¹(-0.4625) ≈ 117.5° (to 1 d.p.)
(b) Area = ½ × 15 × 8 × sin(117.5°) ≈ 53.2 m²
Using the formula for area of triangle with two sides and included angle.
(c) AQ is the hypotenuse of right triangle ACQ.
AQ = √(AC² + CQ²) = √(15² + 4²) ≈ 15.5 m
(d) Vertical difference = CQ – BP = 4 – 3 = 1 m
Horizontal distance = BC = 8 m. Angle = tan⁻¹(1/8) ≈ 7.1°
(e) Midpoint height = (3 + 4)/2 = 3.5 m
First find AQ ≈ 15.5 m from (c). Then angle = tan⁻¹(3.5/15.5) ≈ 11.5°
(a) Calculate angle \( ACD \).
(b) Show that \( BC = 7.05 \, \text{km} \), correct to 2 decimal places.
(c) Calculate the shortest distance from \( B \) to \( AC \).
(d) Calculate the length of the straight line \( BD \).
(e) \( C \) is due east of \( A \). Find the bearing of \( D \) from \( C \).
▶️ Answer/Explanation
(a) 39.6° or 39.57…°
Using the cosine rule: \( \cos ACD = \frac{14^2 + 12^2 – 9^2}{2 \times 14 \times 12} \).
Calculation gives \( \cos ACD ≈ 0.7708 \), then \( ACD ≈ \cos^{-1}(0.7708) ≈ 39.57° \).
(b) 7.05 km (shown)
Using the sine rule: \( \frac{BC}{\sin 25°} = \frac{14}{\sin 123°} \).
Calculate \( BC = \frac{14 \times \sin 25°}{\sin 123°} ≈ 7.054 \) km, which rounds to 7.05 km.
(c) 3.74 km or 3.735 to 3.739 km
The shortest distance is perpendicular from B to AC. Using trigonometry:
Distance = \( BC \times \sin 32° ≈ 7.05 \times \sin 32° ≈ 3.735 \) km.
(d) 11.8 km or 11.83 to 11.85 km
First find angle BCD = angle ACD + 32° ≈ 39.57° + 32° ≈ 71.57°.
Using cosine rule: \( BD^2 = 12^2 + 7.05^2 – 2 \times 12 \times 7.05 \times \cos 71.57° \).
Calculation gives \( BD ≈ 11.83 \) km.
(e) 309.6° or 309.57…°
Bearing is measured clockwise from north. Since C is east of A, angle ACD is 39.57°.
Bearing = 270° + angle ACD ≈ 270° + 39.57° ≈ 309.57°.