Maths

Further trigonometry - Higher

Before starting this section, you should have already revised trigonometry and Pythagoras' theorem.

We are now going to extend trigonometry beyond right-angled triangles and use it to solve problems involving any triangle.

The sine rule

We can use the sine rule to find the size of an angle or length of a side.

The sine rule is:

We can use the sine rule when we are given:

Two sides and an angle opposite to one of the two sides.
One side and any two angles.

Question

Find the size of angle R.

Answer

R = 25.4°

Using the sine we can write:

Multiplying both sides by 4, we get

= 0.4293...

SO R = inv sin(0.4293)

R = 25.4° (1dp)

Remember: Use the formula which has the unknown at the top of the fraction.

Question

Find the length of YZ.

Answer

YZ = 4.40cm?

The angles in a triangle add to 180°.

So angle X was 45°

Now use the formula

The cosine rule

We can use the cosine formula to find the length of a side or size of an angle.

For a triangle with sides a,b and c and angles A, B and C the cosine rule can be written as:

a² = b² + c² - 2bc cos A
or
b² = a² + c² - 2ac cos B
or
c² = a² + b² - 2ab cos C

These formulas can be rearranged to give :

When to use it

We can use the cosine formula when we are given:

Two sides and an angle.
Three sides.

Check whether your formula sheet gives these formulas in both formats. If it does not, you may need to rearrange them yourself.

Question

Find the length of BC.

Answer

Did you get 4.86cm? If so, well done.

If not, remember to use the formula:

a² = b² + c² - 2bc cos A

Substitute the values into the formula.

a² = 7² + 3² - 2 × 3 × 7 cos 35

a² = 58 - 42 cos 35

a² = 23.5956

a = 4.86cm

Question

Find the size of angle R.

Answer

R = 31.8°

You need to use the formula:

R = inv cos 0.8497

Therefore, R = 31.8°

The area of a triangle

Area of triangle ABC = ¹/₂ab sin C

or, ¹/₂ac sin B

or, ¹/₂bc sin A

We can use this formula when we are given two sides and the included angle.

Question

Find the area of triangle ABC.

Answer

Area = 12.4cm²

The area of the triangle is

¹/₂ × 5.2 × 7.1 x sin42 = 12.4cm²

Remember to show all your working. It is easy to make a mistake when using your calculator, and you will not get any marks for a wrong answer.

Finding the angle between a line and a plane

To find the angle between the line PQ and the plane ABCD:

Draw a line from Q which is perpendicular to the plane ABCD and meets the plane at X.
Join X to P.

Click 'Go' in the diagram to show the lines QX and QP:

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We now have a right-angled triangle, so we can use trigonometry. The angle between PQ and the plane ABCD is the angle we have to find.

This method can be applied to all three-dimensional (3-D) problems.

A cuboid

Click 'Go' to reveal the angle between the diagonal AG and the base of the cuboid ABCD:

A square-based pyramid

Click 'Go' to reveal the angle between PV and the base PQRS:

Problems in three dimensions

Try to split problems of this type into a series of two-dimensional questions. It is easier to work in two dimensions than three. Look at the following example.

Question: Calculate the length of VA.

Answer

The pyramid VABCD has a square base that is 4cm long. The height of the pyramid is 3cm.

Using Pythagoras' theorem,

AC² = 4² + 4²

AC² = 32

Therefore, AC =

O is the centre of the base, so AQ is half of AC.

So AO =

Applying Pythagoras' theorem to the triangle AVO, we get

AV² =

AV² = 17

AV = = 4.12cm (3 significant figures (s.f.))

Question: Calculate the angle between VA and the base ABCD.

Answer

The angle between VA and the base ABCD is the angle VÂO.

Using trigonometry

sin VÂO =

= 0.7276

VÂO = 46.7° (1 decimal place)

Try this one yourself:

The triangular prism PQRSTU has a length of 10cm. PTQ and SUR are equilateral triangles with sides that are 4cm long.

Question: Find the perpendicular height of triangle SUR.

Answer

The perpendicular height of triangle SUR is or 3.46cm (3 s.f.).

Let V be the midpoint of SR.

Then VR = 2cm, and UVR is a right-angled triangle.

Using Pythatgoras' theorem:

UV² = UR² - VR²

UV² = 4² - 2²

UV² = 16 - 4

UV² = 12

So the perpendicular height of triangle SUR = or 3.46cm (3 s.f.).

Question: Find the length of QU.

Answer

Applying Pythagoras' theorem to the triangle QUR,

QU² = 10 ² + 4 ²

QU = or 10.8cm (3 s.f.).

Question: Hence, find the angle between QU and the base PQRS.

Answer

QU is part of the triangle QUV, where V is the point we used in the first part of this question, midway between S and R.

For this question, you will need to use the answers from the previous questions. Use the perpendicular height of the prism, UV, which is and the length QU, which is

So in triangle QUV

Sin UQV = /

UQV = Inv sin /

UQV = 18.8° (1dp)

The angle between QU and the base is 18.8° (1 dp).

Remember:

Keep your answers in surd form (like ), or store them in the memory of your calculator. Using rounded values later on in a question will lead to inaccurate answers.

Sine and cosine rules in 3-D

You might also be expected to apply the sine or cosine rules to a 3-D problem. Look at the following example.

Question

Sally and Kate stand some distance away from a building (B) which is 10m high. Sally is on a bearing of 030° from the building. From where she is standing, the angle of elevation of the top of the building is 35°. Kate is on a bearing of 090° from the building. From where she is standing, the angle of elevation of the top of the building is 50°. This information is shown in the diagram.

What is the distance between Sally and Kate?

Answer

Using trigonometry on the right-angled triangles OBS

so = 14.3 m (3 s.f.)

In the same way we can find OK = 8.39m (3 s.f.)

Applying the cosine rule to triangle KOS, we get:

KS² = 8.39² + 14.3² - ( 2 × 8.39 × 14.3 × cos60°)

KS² = 154.534

KS = 12.4m (3 s.f.)

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