Area of a triangle Problem set A Worked solutions

In triangle ABC, AB = 12.4 cm, CA = 8.7 cm and angle A = 57°. Find the area of the triangle.

• The angle 57° is between the sides 12.4 cm and 8.7 cm
• \(\text{Area}=\frac12\times12.4\times8.7\times\sin 57°\)
• From the calculator: area = 45.2 cm²

In triangle PQR, PQ = 6.3 cm, QR = 10.8 cm and angle PQR = 128°. Find the area of the triangle.

• The angle 128° at Q is between the sides 6.3 cm and 10.8 cm
• \(\text{Area}=\frac12\times6.3\times10.8\times\sin 128°\)
• From the calculator: area = 26.8 cm²

Triangle ABC has AB = 11.2 cm, CA = 7.4 cm and area 31.5 cm². Angle A is acute. Find angle A.

• The angle is between the two given sides
• \(31.5=\frac12\times11.2\times7.4\times\sin\theta\)
• \(\sin\theta=\frac{2\times31.5}{11.2\times7.4}\) = 0.7601
• From the calculator: θ = sin⁻¹(0.7601) = 49.5°

Triangle ABC has AB = 10.0 cm, angle A = 55° and area 48 cm². Find the length of CA.

• The angle 55° is between the sides 10.0 cm and x
• \(48=\frac12\times10.0\times x\times\sin 55°\)
• \(x=\frac{2\times48}{10.0\times\sin 55°}\), so CA = 11.7 cm

In triangle ABC, BC = 7.6 cm, CA = 9.4 cm and AB = 13.2 cm. Find the smallest angle.

• The smallest angle is opposite the shortest side, BC: it is angle A
• \(\cos\theta=\frac{9.4^2+13.2^2-7.6^2}{2\times9.4\times13.2}\) = 0.8254
• θ = cos⁻¹(0.8254) = 34.4°

In triangle PQR, the perpendicular from Q meets PR at S and QS = h. The sides are p = QR and q = RP. Show that the area of the triangle is \(\frac12 pq\sin R\).

• In triangle RSQ: \(\sin R=\frac{h}{p}\), so \(h=p\sin R\)
• The base is RP = q and the height is h: \(\text{Area}=\frac12\times q\times h\)
• Substitute: \(\text{Area}=\frac12\times q\times p\sin R=\frac12 pq\sin R\)

A triangular sail has two edges of 4.8 m and 3.9 m. The angle between them is 72°. Find the area of the sail.

• The angle 72° is between the sides 4.8 m and 3.9 m
• \(\text{Area}=\frac12\times4.8\times3.9\times\sin 72°\)
• From the calculator: area = 8.90 m²

A triangular roof panel RST has area 7.5 m². RS = 3.4 m and angle R = 58°. Find the length of TR.

• The angle 58° is between the sides 3.4 m and x
• \(7.5=\frac12\times3.4\times x\times\sin 58°\)
• \(x=\frac{2\times7.5}{3.4\times\sin 58°}\), so TR = 5.20 m

A triangular plot of land has sides of 62 m, 85 m and 104 m.

1. (a) Find the largest angle of the plot.

   • \(\cos\theta=\frac{62^2+85^2-104^2}{2\times62\times85}\) = 0.02400
   • θ = cos⁻¹(0.02400) = 88.6°

2. (b) Find the area of the plot.

   • θ is between the sides 62 m and 85 m
   • \(\text{Area}=\frac12\times62\times85\times\sin 88.6°\) = 2630 m²

Triangle ABC has AB = 9.0 cm, CA = 7.0 cm and area 24 cm². Find the two possible sizes of angle A.

• \(24=\frac12\times9.0\times7.0\times\sin A\)
• \(\sin A=\frac{2\times24}{9.0\times7.0}\) = 0.7619
• From the calculator: A = 49.6°
• A sine is also positive for an obtuse angle: A = 180° − 49.6° = 130°

In the quadrilateral ABCD, AB = 7.3 cm, BC = 9.6 cm, CD = 8.8 cm, DA = 6.1 cm and angle ABC = 64°.

1. (a) Find the area of triangle ABC.

   • \(\text{Area}=\frac12\times7.3\times9.6\times\sin 64°\) = 31.5 cm²

2. (b) Find the length of AC.

   • \(AC^2=7.3^2+9.6^2-2\times7.3\times9.6\times\cos 64°\) = 84.01
   • AC = 9.17 cm

3. (c) Find angle ADC.

   • \(\cos\theta=\frac{6.1^2+8.8^2-9.166^2}{2\times6.1\times8.8}\) = 0.2854
   • θ = cos⁻¹(0.2854) = 73.4°

4. (d) Find the area of the quadrilateral ABCD.

   • Area of triangle ADC = \(\frac12\times6.1\times8.8\times\sin 73.42°\) = 25.7 cm²
   • Total = 31.49 + 25.72 = 57.2 cm²