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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**.

- Answer
The pyramid

**VABCD**has a square base that is 4cm long. The height of the pyramid is 3cm.Using Pythagoras' theorem,

**AC**^{2}= 4^{2}+ 4^{2}**AC**^{2}= 32Therefore,

**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**AV**^{2}=**AV**^{2}=**AV**^{2}= 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:

**UV**^{2}=**UR**^{2}-**VR**^{2}**UV**^{2}= 4^{2}- 2^{2}**UV**^{2}= 16 - 4**UV**^{2}= 12So 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**,**QU**^{2}= 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.

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:**KS**^{2}= 8.39^{2}+ 14.3^{2}- ( 2 × 8.39 × 14.3 × cos60°)**KS**^{2}= 154.534**KS**= 12.4m (3 s.f.)

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