Unit 15 Section 4 : Calculating Angles

In this section we use trigonometry to determine the sizes of angles in rightangled triangles. On your scientific calculator you will find buttons labelled 'sin–1' , 'cos–1' and 'tan–1' . You will need to be able to use these to calculate the angles that will arise in the problems which follow. Again, we start with the three trigonometric functions:

Trigonometric Functions
sinΘ  =  cosΘ  =  tanΘ  = 

Example 1

Calculate the angle Θ in this triangle.

In this triangle we are given the lengths of the adjacent and opposite sides, so we will use,

tanΘ =

Using the lengths given, we have

tanΘ =
= 1.6

We can then use the tan–1 key on a calculator to obtain

Θ = tan–1(1.6) = 57.99461678°
= 58.0° (to 1 decimal place)

Example 2

Calculate the angle marked Θ in this triangle.

Because the lengths given are for the adjacent side and the hypotenuse, the formula for cosΘ must be used.

cos Θ =
= = 0.470588235
Θ = cos–1(0.470588235) = 61.92751306°
= 61.9° (to 1 decimal place)

Example 3

A rectangle has sides of length 5 m and 10 m. Determine the angle between the long side of the rectangle and a diagonal.

The solution is illustrated in the diagram.

Using the formula for tanΘ gives

tan Θ =
= 0.5

Then using the tan–1 key on a calculator gives

Θ = tan–1(0.5) = 26.56505118°
= 26.6° (to 1 decimal place)

Exercises

Question 1

Giving your answers, where necessary, correct to 1 decimal place, use your calculator to obtain Θ if:

(a)sin Θ = 0.8 Θ = °
(b)cos Θ = 0.5 Θ = °
(c)tan Θ = 1 Θ = °
(d)sin Θ = 0.3 Θ = °
(e)cos Θ = 0 Θ = °
(f)tan Θ = 14 Θ = °
Question 2

Use the tangent function to calculate the angle Θ in each of the following diagrams. In each case, give your answer correct to 1 decimal place.

(a)
Θ = °
(b)
Θ = °
(c)
Θ = °
(d)
Θ = °
Question 3

Use sine or cosine to calculate the angle Θ in each of the following triangles. In each case, give your answer correct to 1 decimal place.

(a)
Θ = °
(b)
Θ = °
(c)
Θ = °
(d)
Θ = °
Question 4

Calculate the angle Θ in each of the following triangles. In each case, give your answer correct to 1 decimal place.

(a)
Θ = °
(b)
Θ = °
(c)
Θ = °
(d)
Θ = °
Question 5

A right-angled triangle has sides of length 3 cm, 4 cm and 5 cm.

Determine the sizes of all the angles in the triangle, giving your answers to the nearest degree.

°, ° and °
Question 6

The diagram shows the cross-section of a shed.

Calculate the angle Θ between the roof and the horizontal. Give your answer to the nearest degree.

Θ = °
Question 7

A ladder of length 6 m leans against a wall. The foot of the ladder is at a distance of 3 m from the base of the wall.

Calculate the angle between the ladder and the ground.

Angle with ground = °
Question 8

A rectangle has sides of length 12 cm and 18 cm.

(a)

Calculate the length of the diagonal of the rectangle, giving your answer to the nearest millimetre.

Length of diagonal = cm

L2 = 122 + 182 = 144 + 324 = 468

L = = 21.63330765 cm = 21.6 cm (to the nearest mm)

(b)

Calculate the angle between the diagonal and the shorter side of the rectangle, giving your answer to the nearest degree.

Angle between diagonal and shorter side = °
A = tan–1(1.5) = 56.30993247° = 56° (to the nearest degree)
Question 9

As an aeroplane travels 3000 m along a straight flight path, it rises 500 m.

Calculate the angle between the flight path of the aeroplane and the horizontal. Give your answer to a sensible level of accuracy.

Angle between flight path and horizontal = °
Question 10

A weight hangs from 2 strings as shown in the diagram.

Calculate the angle between the two strings, giving your answer to the nearest degree.

Angle between strings = °
A = cos–1(0.9) + cos–1(0.75) = 25.84193276° + 41.40962211° = 67.25155487 = 67° (to the nearest degree)
Question 11

Ramps help people going into buildings.
A ramp that is 10 m long must not have a height greater than 0.83 m.

(a)

Here are the plans for a ramp:

Is this ramp too high?

Height of top of ramp = m (to the nearest cm).

So the ramp is .

H2 = 102 – 9.852 = 100 – 97.0225

H = = 1.725543393 m

1.73 m > 0.83 m, so the ramp is too high

(b)

Here are the plans for a different ramp.

How long is the base of this ramp?

m (to the nearest cm)
L = 10 × cos3.5° = 9.981347984 m = 9.98 m (to the nearest cm)
(c)

The recommended gradient of a ramp is 1 in 20.

What angle gives the recommended gradient?

° (to the nearest degree)
Recommended angle = tan–1 = tan–1(0.05) = 2.862405226° = 3° (to the nearest degree)
Question 12

A boat sails from the harbour to the buoy.
The buoy is 6 km to the east and 4 km to the north of the harbour.

(a)

Calculate the shortest distance between the buoy and the harbour. Give your answer to 1 decimal place.

Shortest distance = km

SD2 = 62 + 42 = 36 + 16 = 52

SD = = 7.211102551 km

(b)

Calculate the bearing of the buoy from the harbour.

Bearing = ° (to the nearest degree)
Angle between direction and north = tan–1(1.5) = 56.30993247°

The buoy is 1.2 km to the north of the lighthouse. The shortest distance from the lighthouse to the buoy is 2.5 km.

(c)

Calculate how far the buoy is to the west of the lighthouse. Give your answer to 1 decimal place.

Distance west = km

DW2 = 2.52 – 1.22 = 6.25 – 1.44 = 4.81

DW = = 2.19317122 km

Question 13

Bargate is 6 km east and 4 km north of Cape Point.

(a)

Steve wants to sail directly from Cape Point to Bargate.

On what bearing should he sail?

Bearing = ° (to the nearest degree)
Angle between direction and north = tan–1(1.5) = 56.30993247°
(b)

Anna sails from Cape Point on a bearing of 048 °. She stops when she is due north of Bargate.

How far north of Bargate is Anna?

Distance north of Bargate = km (to 2 s.f.)

Distance north of Cape Point = 6 ÷ tan48° = 5.402424266 km

Distance north of Bargate = 5.402424266 – 4 = 1.402424266 km