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Further trigonometry - Higher

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# 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.

VABCD is a square based pyramid.

Question

Calculate the length of VA.

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 =

AV2 = 17

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

Question

Calculate the angle between VA and the base ABCD.

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.

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.

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.

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?

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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