At the end of the lesson, students should be able to;

- Demonstrate a clear understanding of trigonometry in three dimensions.
- Use trigonometry to solve three dimensional problems.

__The plane__

A plane is a surface such as the top of a table or the cover of a book.

__The angle between a line and a plane__

In the figure below, the line PA intersects the plane WXYZ at A. To find the angle between PA and the plane, draw PL which is perpendicular to the plane and join AL. The angle between PA and the plane is PAL.

__The angle between two planes__

Two planes which are not parallel intersect in a straight line. Examples of this are the floor and a wall of a room and two walls of a room. To find the angle between two planes draw a line in each plane which is perpendicular to the common line of intersection. The angle between the two lines is the same as the angle between the two planes.

Three planes usually intersect at a point as, for instance, two walls and the floor of a room.

Problems with solid figures are solved by choosing suitable right-angled triangles in different planes. It is essential to make a clear three-dimensional drawing in order to find these triangles. The examples which follow show the methods that should be adopted.

__Example 1 __

The figure below is a cuboid. Calculate the length of the diagonal AG.

__Solution__

The figure below shows that in order to find AG we must use the right-angled triangle AGE. EG is the diagonal of the base rectangle.

In triangle EFG, EF = 8 cm, GF = 6 cm and angle EFG = 90^{0}

Using Pythagoras’ theorem,

**Always draw a large, clear diagram. It is always helpful to redraw the triangle which contains the length or angle to be found.**

**To find the value of an angle or side in a three dimensional figure you need to find a right – angled triangle in the figure which contains it. This triangle also has to contain two known values that you can use in the calculation.**

**You must redraw this triangle separately as a plain, right – angled triangle. Add the known values and the unknown values you want to find. Then use the trigonometric ratios and Pythagoras theorem to solve the problem. **

__Example 2__

X, Y and Z are three points at a ground level. They are in the same horizontal plane. Z is 50km east of Y. Y is in the northern part of X. Z is on a bearing of 050ᵒ from X. An aircraft, flying east, passes over Y and over Z at the same height. When it passes over Y, the angle of elevation from X is 12ᵒ. Find the angle of elevation of the aircraft from X when it is over Z.

__Solution__

First, draw a diagram containing all the known information.

**The Angle between a Line and a Plane**

To calculate the size of the angle between the line AB and the standard plane, drop a perpendicular from B. It meets the shaded plane at C. Then join AC.

The angle between the lines AB and AC represents the angle between the line AB and the shaded plane. The line AC is the projection of the line AB on the shaded plane.

__Example 3__

A rectangular box with top WXYZ and base ABCD has AB = 6cm, BC = 8cm and WA = 3cm. Calculate

a) The length of AC

b) The angle between WC and AC

__Solution__

a) Redraw triangle ABC.

b) Redraw triangle WAC

__The sine and cosine rule in three dimensions__

The sine and cosine rules can be used to find lengths and angles in three dimensional situations. In all cases use a two-dimensional triangular section.

__Example 4__

A tower 100 m high stands in a flat plane. From the top of the tower, village A is on a bearing of 083^{0}, and the angle of depression is 5^{0}. Village B is on a bearing of 118^{0}, and the angle of depression is 4^{0}. Find the distance between the villages.

__Solution __

We know two sides of triangle ABC and the included angle. Use the cosine rule to find the third side.

AB^{2} = 1143^{2} + 1430^{2} – 2 × 1143 × 1430 × cos 35^{0} = 673600

The distance between the villages is 821 m.

__Note __

Always write down intermediate working values to at least 4 significant figures, or use the answer on your calculator display to avoid inaccuracy in the final answer.

__Unit 10 lesson 20: Exercise__

1) The diagram below is a cuboid with a square base HEFG of side 3 cm.

a) Calculate the length CE

b) Calculate the angle GFH

The diagram shows a right pyramid where U is vertically above X.

a) Calculate the length WY

b) Calculate the length UX

c) Calculate the angle between UX and UZ.

3)

Using the triangular prism, calculate:

a) The length AD

b) The length AC

c) The angle AC makes with the plane CDEF.

d) The angle AC makes with the plane ABFE.

4) The diagram shows a triangular pyramid on a horizontal base ABC, V is vertically above B where VB = 10cm, <ABC = 90ᵒ and AB = BC = 15cm. Point M is the mid – point of AC. Calculate the size of angle VMB.

5) The angle of elevation of the top of a tower is 38ᵒ from a point A due south of it. The angle of elevation of the top of the tower from another point B, due east of the tower is 29ᵒ. Find the height of the tower if the distance AB is 50m.

6) An observer at the top of a tower of height 15m sees a man due west of him at an angle of depression 31ᵒ. He sees another man due south at an angle of depression 17ᵒ. Find the distance between the men.

7) The angle of elevation of the top of a tower is 27ᵒ from a point A due east of it. The angle of elevation of the top of the tower is 11ᵒ from another point B due south of the tower. Find the height of the tower if the distance AB is 40m.

8)

A cliff is 200 m high. From the top of the cliff two ships are seen: one is at an angle of depression of 3.7^{0} and on a bearing of 352^{0}, the other is at an angle of depression of 2.6^{0} and a bearing of 048^{0}. Find the distance between the ships.

9) The angle of elevation of the top of a tower is 27ᵒ from a point A due east of it. The angle of elevation of the top of the tower is 11ᵒ from another point B due south of the tower. Find the height of the tower if the distance AB is 40km.

10) An aircraft is vertically above a point which is 10 km West and 15 km North of a control tower. If the aircraft is 4000 m above the ground, how far is it from the control tower?

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