Applications of trigonometry for SQA National 5 Maths

Applications of trigonometry

This page covers the following topics:

1. Area of an irregular triangle
2. Sine rule
3. Cosine rule

The area of any triangle can be found when an angle and the lengths of the two sides subtending it are known. Given that C is the angle and the sides are a and b, the area of the triangle is given by 1/2 × absinC.

Area of an irregular triangle

The sine rule states that the ratio of the sine value of an angle and the length of the side opposite it is equal for all sides of a triangle. This is given as sinA/a = sinB/b = sinC/c, which is equivalent to a/sinA = b/sinB = c/sinC. The first version is usually used to find angles, whereas the second version is usually used to find lengths of sides.

Sine rule

In a triangle with a side a opposite an angle A and two other sides b and c, the cosine rules states that: a² = b² + c² − 2bccosA. This can be rearranged to cosA = (b² + c² − a²) ÷ 2bc to find angles more easily.

Cosine rule

1

A triangle has an angle of size 40° and an angle of 70°. Given that the sides subtending the unknown angle have lengths 8 cm and 10 cm, calculate the area of the triangle.

The sum of the angles in a triangle is 180°.
40° + 70° + θ = 180°
θ = 180° − 70° − 40°
θ = 70°

Area = 1/2 × absinC
Area = 1/2 × 8 cm × 10 cm × sin(70°)
Area = 37.6 cm² (to 3 s. f.)

37.6 cm²

A triangle has an angle of size 40° and an angle of 70°. Given that the sides subtending the unknown angle have lengths 8 cm and 10 cm, calculate the area of the triangle.

2

Zain is making a garland out of triangles. Given that each triangle has two sides of 8 cm and 10 cm, and an angle of 60° opposite a side of unknown length, calculate the length of the third side.

By the cosine rule, a² = b² + c² − 2bccosA.
a² = 10² + 8² − 2 × 10 × 8 × cos(60°)
a² = 84
a = √74.4 = 2√21 cm

2√21 cm

Zain is making a garland out of triangles. Given that each triangle has two sides of 8 cm and 10 cm, and an angle of 60° opposite a side of unknown length, calculate the length of the third side.

3

Find the area of the given triangle.

By the cosine rule, cosA = (b² + c² − a²) ÷ 2bc.
cosA = (8² + 9² − 12²) ÷ (2 × 8 × 9)
cosA = 1/144
a = arccos(1/144) = 89.6° (to 3 s. f.)

Area = 1/2 × absinC
Area = 1/2 × 8 cm × 9 cm × sin(89.6°)
Area = 36.0 cm² (to 3 s. f.)

36.0 cm²

Find the area of the given triangle.

4

An angle of size 45° is subtended by a side of length 6 cm and a side of unknown length in a triangle. Given that the area of the triangle is 3√2 cm², find the unknown length.

Let x be the unknown length.
Area = 1/2 × absinC
3√2 cm² = 1/2 × 6 cm × x × sin(45)
x = 2 cm

2 cm

An angle of size 45° is subtended by a side of length 6 cm and a side of unknown length in a triangle. Given that the area of the triangle is 3√2 cm², find the unknown length.

5

Calculate the perimeter of the given triangle, given that it has an area of 20 cm².

Area = 1/2 × absinC
20 cm² = 1/2 × 5 cm × 10 cm × sinC
sinC = 4/5
C = arcsin(4/5) = 53.1°

By the sine rule, a/sinA = b/sinB.
a/sin(53.1°) = 5 cm/sin(30°)
a = sin(53.1°) × 5 cm ÷ sin(30°)
a = 8 cm

Perimeter = 8 cm + 10 cm + 5 cm = 23 cm

23 cm

Calculate the perimeter of the given triangle, given that it has an area of 20 cm².

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