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Maths

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.

image: pyramid

 

VABCD is a square based pyramid.

Question

Calculate the length of VA.

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Answer

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

image: one square or two triangles

Using Pythagoras' theorem,

AC2 = 42 + 42

AC2 = 32

Therefore, AC = square root of 32

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

So AO = square root of 32 divided by 2

image:triangle

Applying Pythagoras' theorem to the triangle AVO, we get

AV2 = (square root of 32 over 2) to the power of 2 + 3 to the power of 2

AV2 = 32 over 4 + 9

AV2 = 17

AV = square root of 17 = 4.12cm (3 significant figures (s.f.))

Question

Calculate the angle between VA and the base ABCD.

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Answer

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

image: triangle

Using trigonometry

sin VÂO = 3 over square root of 17

3 over square root of 17 = 0.7276

VÂO = 46.7° (1 decimal place)

Try this one yourself:

image: prism

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.

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Answer

The perpendicular height of triangle SUR is square root of 12 or 3.46cm (3 s.f.).

image: triangle

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 = square root of 12 or 3.46cm (3 s.f.).

Question

Find the length of QU.

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Answer

Applying Pythagoras' theorem to the triangle QUR,

QU2 = 10 2 + 4 2

QU = square root of 116 or 10.8cm (3 s.f.).

Question

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

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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 square root of 12 and the length QU, which is square root of 116

So in triangle QUV

Sin UQV = square root of 12 / square root of 116

UQV = Inv sinsquare root of 12 / square root of 116

UQV = 18.8° (1dp)

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

Remember:

Keep your answers in surd form (like square root of 12), 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.

image: pyramid

What is the distance between Sally and Kate?

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Answer

Using trigonometry on the right-angled triangles OBS

10 over OS = tan 35 degrees so OS = 10 over tan 35 degrees = 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.)

image: triangle scalene

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