Key points about 2D and 3D trigonometry problems
Trigonometry at Higher tier may combine other parts of geometry. This could include problems that involve:
- compound shapes
- areas of triangles using trigonometry
- bearings
- angles between lines and planes in 3D shapes
Being able to identify which formula to use, sine or cosine, and when, is necessary to answer questions on this topic successfully.
Check you are confident in how to use the sine and cosine rules.
How to find the area of a triangle using the trigonometric formula
An alternative method for finding the area of a triangle is used, when two sides and the An angle between two given sides. are known.
Calculate the area of a triangle using the following A fact, rule, or principle that is expressed in terms of mathematical symbols. The plural of formula is formulae. :
Area of a triangle = \(\frac{1}{2}\)𝑎𝑏 sin𝐶
In the formula 𝑎 and 𝑏 are the lengths of the two known sides and 𝐶 is the size of the angle between them.
Follow the worked example below
GCSE exam-style questions
- Calculate the area of triangle 𝑃𝑄𝑅.
Give the answer to 1 decimal place.
Area = 36·8 cm² to 1 decimal place.
- Label the sides of the triangle. Since the vertices are not called 𝐴, 𝐵 and 𝐶 let vertex, 𝑃, with angle 130° be 𝐶.
The choice of 𝐴 and 𝐵 does not matter. Let vertex 𝑄 be 𝐴 and vertex 𝑅 be 𝐵.
The 12 cm side, opposite angle 𝐴, is called 𝑎. The 8 cm side, opposite angle 𝐵, is called 𝑏.
- Substitute the values of 𝑎, 𝑏 and 𝐶 into the formula to give
Area = \(\frac{1}{2}\) × 12 × 8 × sin(130)
- Type \(\frac{1}{2}\) × 12 × 8 × sin(130) into a scientific calculator.
Usually, the calculator will automatically open a bracket after pressing the sine button.
Remember to close the bracket after typing in the angle.
- This gives Area = 36·7701…
Therefore, rounded to one decimal place, Area = 36·8 cm².
- Calculate the area of the parallelogram.
Give the answer to 1 decimal place.
Area = 63·6 m² to 1 decimal place.
Divide the parallelogram in two by adding a diagonal. The area of the parallelogram is twice the area of the triangle.
Label the sides of the triangle.
Since the vertices are labelled, let angle 130° be 𝐶.
The choice of 𝐴 and 𝐵 does not matter.
The 5 m side, opposite angle 𝐴, is called 𝑎.
The 13 m side, opposite angle 𝐵, is called 𝑏.
- Substitute the values of 𝑎, 𝑏 and 𝐶 into the formula to give
Area = \(\frac{1}{2}\) × 12 × 8 × sin(130)
- Type \(\frac{1}{2}\) × 5 × 13 × sin(78) into a scientific calculator.
Usually, the calculator will automatically open a bracket after pressing the sine button.
Remember to close the bracket after typing in the angle.
This gives the area of the triangle = 31·7897…
- Find the area of the parallelogram by multiplying this by two.
31·7897… × 2 = 63·5795…
Therefore, rounded to one decimal place, the area of the parallelogram = 63·6 m².
- The area of the triangle is 30 cm².
Calculate the size of the acute angle 𝑥 to 1 decimal place.
𝑥 = 59·0° to 1 decimal place.
Label the sides of the triangle. The 10 cm side, opposite angle 𝐴, is called 𝑎. The 7 cm side, opposite angle 𝐵, is called 𝑏.
Substitute the values of 𝑎, 𝑏, 𝐶 and the area into the formula to give 30 = \(\frac{1}{2}\) × 10 × 7 × sin(𝑥)
\(\frac{1}{2}\) × 7 × 10 = 35. This simplifies to 30 = 35 × sin(𝑥).
Find sin(𝑥) by dividing both sides by 35. This gives 30 ÷ 35 = sin(𝑥).
Work out angle, 𝑥, by using the inverse function of sine.
𝑥 = sin⁻¹ (30/35).
- To write sin⁻¹ on a scientific calculator, press 'Shift' then 'sin'. Then type 30 ÷ 35. Remember to close the brackets.
This gives 𝑥 = 58·9972…
Rounded to 1 decimal place, 𝑥 = 59·0°.
How to apply the sine and cosine rules
To solve problems involving non-right-angled triangles, the correct formula must be applied.
Use the sine rule on any triangle to calculate:
- a side when two angles and an opposite side are known, eg 𝑎, 𝐴 and 𝐵
- an angle when two sides and an opposite angle are known, eg 𝑎, 𝐴 and 𝑏
Use the cosine rule on any triangle to calculate:
- a side when two sides and the included angle are known, eg 𝑎, 𝑏 and 𝐶
- an angle when all three sides are known
- Always add any extra angles you can calculate to a diagram, using the rules of geometry.
Follow the worked example below
GCSE exam-style questions
- An oil tanker sails from 𝐷 to 𝐸 and then from 𝐸 to 𝐹.
𝐸 is 25 miles from 𝐷, on a bearing of 050°.
𝐹 is 32 miles from 𝐸, on a bearing of 105°.
Work out the direct distance from 𝐷 to 𝐹.
𝐷𝐹 = 50·7 miles, to one decimal place.
- First calculate the other angles using geometry. The two north lines are parallel.
Co-interior angles add up to 180°. The angle to the left of 𝐸 = 180 – 50 = 130°.
Use angles at a point to calculate angle 𝐷𝐸𝐹.
Angle 𝐷𝐸𝐹 = 360 – 105 – 130 = 125°.
In the triangle two sides and the included angle are known. The cosine rule is used to work out the missing side.
- Label the sides of the triangle.
Here the vertices are not labelled using 𝐴, 𝐵 and 𝐶, so rename the vertex with angle 125° as 𝐴.
The choice of 𝐵 and 𝑏 does not matter. Let 𝐷 be 𝐵 and 𝐹 be 𝐶.
The 25-mile side, opposite angle 𝐶, is called 𝒄.
The 32-mile side, opposite angle 𝐵, is called 𝑏.
The side 𝐵𝑏, opposite angle 𝐴, is called 𝑎.
- Substitute the values of 𝐴, 𝑎, 𝑏 and 𝑐 into the formula to give 𝐷𝐹² = 32² + 25² – (2 × 32 × 25) cos(125).
32² = 1024 , 25² = 625 and 2 × 32 × 25 = 1600, so this simplifies to 𝐷𝐹² = 1024 + 625 – 1600cos(125).
- Type 1024 + 625 – 1600cos(125) into a scientific calculator.
Usually, the calculator will automatically open a bracket after pressing the cosine button.
Remember to close the bracket after typing in the angle. This gives 𝐷𝐹² = 2566·7222…
It is important not to round the number at this stage.
- Find 𝐷𝐹 by calculating the square root of 2566·7222…
Type the square root button followed by the ‘Ans’ button into a scientific calculator.
This gives the answer of 𝐷𝐹 = 50·6628…. Therefore, rounded to one decimal place, 𝐷𝐹 = 50·7 miles.
- Quadrilateral 𝑊𝑋𝑌𝑍 is formed by combining two triangles.
Work out the length of 𝑊𝑋. Give the answer to one decimal place.
𝑊𝑋 = 13·4 cm, to one decimal place.
Split the quadrilateral into two separate triangles and apply non-right angled trigonometry.
Step 1. Calculate the length of 𝑋𝑍 using the sine rule.
- Label the sides of the triangle. Since the vertices are not called 𝐴, 𝐵 and 𝐶 let vertex 𝑍 be 𝐴, 𝑋 be 𝐵 and 𝑌 be 𝐶₁.
The 12 cm side, opposite angle 𝐵, is called 𝑏. The side 𝑋𝑍 or 𝐴𝐵, opposite angle 𝐶, is called 𝒄, and the unknown side, opposite angle 𝐴, is called 𝑎.
Since neither side 𝑎, or angle 𝐴 is known this is the portion of the sine rule formula that will not be used.
Substitute the values of 𝐵, 𝐶, 𝑏 and 𝑐 into the formula to give \(\frac{12}{sin(66)}\) = \(\frac{𝐴𝐵}{sin(97)}\).
Rearrange the equation to make 𝐴𝐵 the subject. Find the value of 𝐴𝐵 by multiplying both sides of the equation by sin(97). This gives \(\frac{12sin(97)}{sin(66)}\) = 𝐴𝐵.
Type 12sin(97) ÷ sin(66) into a scientific calculator. This gives 𝐴𝐵 = 13·0377 … It is important not to round the number at this stage.
Step 2. Calculate the length of 𝑊𝑋 using the cosine rule.
- Label the sides of the triangle. Keep 𝑍 as 𝐴, 𝑋 as 𝐵 and let 𝑊 be 𝐶₂.
The 9 cm side, opposite angle 𝐵, is called 𝑏. The side, opposite angle 𝐶₂, is called 𝑐, and is the answer that was calculated in Step 1.
The side labelled 𝑊𝑋, opposite angle 𝐴, is called 𝑎.
- Substitute the values of 𝐴, 𝑎, 𝑏 and 𝑐 into the formula to give 𝑊𝑋² = 92 + 𝑋𝑍² - (2 × 10 × 𝑋𝑍) cos(72).
92 = 81 and 2 × 9 × 𝑋𝑍 = 18𝑋𝑍, so this simplifies to 𝑎² = 81+ 𝑋𝑍² – 18𝑋𝑍cos(72).
- Type 81 + 𝑋𝑍² – 18𝑋𝑍cos(72) into a scientific calculator. Use the ‘Ans’ button each time you need 𝑋𝑍.
This gives 𝑊𝑋² = 178·4642…
It is again important not to round the number at this stage.
- Find 𝑊𝑋, by calculating the square root of 178·4642…
Type the square root button followed by the ‘Ans’ button into a scientific calculator.
This gives the answer of 𝑊𝑋 = 13·3589….
Therefore, rounded to one decimal place, 𝑊𝑋 = 13·4 cm.
How to find the angle between a line and a plane
The angle between a line and a A two-dimensional surface. is the smallest angle between the line and its A correspondence between points of one figure and a second surface. onto the plane.
The projection is a line which can be found by dropping a vertical from the end of the line to the plane and joining it to the other end of the line. These lines form a right-angled triangle.
The angle is calculated using right-angled trigonometry. Pythagoras’ theorem may also be required.
Follow the worked example below
GCSE exam-style questions
- Calculate the size of angle 𝐵𝐷𝐸.
Give the answer to one decimal place.
Angle 𝐵𝐷𝐸 = 19·1°, to one decimal place.
First identify the angle to be calculated.
Angle 𝐵𝐷𝐸, marked as θ, is the angle between 𝐸𝐷 and the plane 𝐴𝐵𝐶𝐷.
To calculate angle 𝐵𝐷𝐸 two sides of the triangle must be known.
The lengths of 𝐴𝐹 and 𝐵𝐸 are equal, so 𝐵𝐸 = 5 cm.
Step 1. Calculate the length of 𝐵𝐷 using Pythagoras’ theorem.
For triangle 𝐵𝐶𝐷, Pythagoras’ theorem states 𝐶𝐷² + 𝐵𝐶² = 𝐵𝐷². Substitute the values, 𝐶𝐷² = 8² and 𝐵𝐶² = 12², into the formula.
Calculate the value of the squares. 8² = 64, and 12² = 144. Add the squares together to get 208. This is the value of 𝐵𝐷².
Find 𝐵𝐷 by calculating the square root of 208. Expressed as a surd 𝐵𝐷 = √208.
Step 2. Now two sides of the triangle 𝐵𝐷𝐸 are known, calculate angle 𝐵𝐷𝐸 using right-angled trigonometry.
In this triangle, the opposite (opp) and the adjacent (adj) are known. The trigonometric ratio needed must contain the opposite and the adjacent. The correct formula to use is the tangent ratio, tanθ = opp ÷ adj.
Substitute the values of opp, adj and θ into the formula to form an equation. Here the opposite is 5, the adjacent is √208 and the angle should be substituted with θ.This gives tanθ = 5 ÷ √208.
Work out the angle, θ, using the inverse function of tangent.
Press ‘Shift' then 'tan' to write tan⁻¹ on a scientific calculator.
Then type 5 ÷ √208.
Remember to close the brackets.
This gives θ = 19·1207… Rounded to 1 decimal place, angle 𝐵𝐷𝐸 = 19·1°.
- Work out the angle between the line 𝐷𝐹 and the plane 𝐸𝐹𝐺𝐻.
Give the answer to one decimal place.
θ = 28·3°, to one decimal place.
First identify the angle to be calculated.
Drop a vertical line from 𝐷 to the plane. It meets the plane at point 𝐻. The line from 𝐻 to 𝐹 is the projection of line 𝐷𝐹. These lines form a right-angled triangle.
The angle to be calculated is θ. In three letter notation this is angle 𝐷𝐹𝐻. To calculate angle θ, two sides of the triangle must be known.
Step 1. Calculate the length of 𝐹𝐻 using Pythagoras’ theorem.
For triangle 𝐹𝐺𝐻, Pythagoras’ theorem states 𝐹𝐺² + 𝐺𝐻² = 𝐹𝐻². The lengths of 𝐹𝐺 and 𝐴𝐷 are equal, so 𝐹𝐺 = 5 cm. Substitute the values, 𝐹𝐺² = 5² and 𝐺𝐻² = 12², into the formula.
Calculate the value of the squares. 5² = 25, and 12² = 144. Add the squares together to get 169. This is the value of 𝐹𝐻².
Find 𝐹𝐻, by calculating the square root of 169. 𝐹𝐻 = √169 = 13 cm.
Step 2. Now two sides of the triangle 𝐷𝐹𝐻 are known, calculate angle θ, using right-angled trigonometry.
In this triangle, the opposite (opp) and the adjacent (adj) are known. The trigonometric ratio needed must contain the opposite and the adjacent.
The correct formula to use is the tangent ratio, tanθ = opp ÷ adj.
Substitute the values of opp, adj and θ into the formula to form an equation. Here the opposite is 7, the adjacent is 13 and the angle should be substituted with θ. This gives tanθ = 7 ÷ 13
Work out the angle, θ, using the inverse function of tangent.
Press 'Shift' then 'tan' to write tan⁻¹ on a scientific calculator.
Then type 7 ÷ 13.
Remember to close the brackets.
This gives θ = 28·3007… Rounded to 1 decimal place, θ = 28·3°.
Check your understanding
Quiz – 2D and 3D trigonometry problems
Practise what you know about 2D and 3D trigonometry problems with this quiz.
Now you've revised 2D and 3D trigonometry problems, why not look at right-angled trigonometry?
More on Geometry and measure
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