The sides of any right angled triangle can be labelled

Hypotenuse, Adjacent and Opposite.

The hypotenuse is always opposite the right angle.

It is the longest side of the triangle.

The adjacent is the side that forms part of the required angle.

The opposite is the side directly across from the required angle.

or from the top angle,

The ratio of these sides is given special names:

sine, cosine and tangent.

These are shortened to sin, cos and tan

To find an angle or side, follow this recipe:-

- Find and sketch the triangle
- Mark the right angle
- Identify and mark angle to be used / found
- Label Opposite, Hypotenuse and Adjacent
- Write out ratio
- Write down solution

* Always draw a sketch*

Example

An aircraft is 73m above a building and 200m from its touchdown point.

Calculate angle θ˚, the glide path of the aircraft.

Give your answer correct to one decimal place.

The angle of the glide path of the aircraft is 20.1˚ (1dp)

Example

An aircraft is 200m from its touchdown point, on a glide path of 30˚.

How high above the ground is the aircraft?

Give your answer correct to one decimal place.

Example

A skier is racing down a 150m long ramp which has an angle of inclination of 30°.

How high above the ground is the starting flag?

The starting flag is 50m above ground.

Example

A skier is racing down a ramp which is 200m long.

The start of the ramp is 100m above ground level.

What is the value of θ˚, the angle of inclination of the ramp?

The angle of inclination of the ramp is 30˚.

Example

A ship is at anchor. The chain is 150 m long and makes an angle of 30˚ from the anchor point to the seabed. The anchor point is 5m above the water level. How deep in the water is the anchor ? Give your answer to the nearest metre.

Example

A ship is taut at anchor. The chain is 150 m long and lies 75m from the anchor point. What is the value of θ˚, the angle from the seabed to the anchor point at the ship’s bow ?

The following recipe can be used for trig problem solving questions :-

- Find and sketch the triangle
- Mark the right angle
- Identify and mark angle to be used / found
- Label Opposite, Hypotenuse and Adjacent
- Write out

- Two Tick Test:-

Tick what you know

Tick what you want

Look for two ticks and use that ratio - Write out ratio and solve.
- Write down solution

Example

How high up the wall is the ladder ?

(Give your answer to 2d.p.)

Tick what you know Adjacent

Tick what you want Opposite

Look for two ticks and use that ratio.

( Here, use tan)

The ladder is 1.73m (2dp) up the wall.

What is the area of this triangle?

Method 1

Draw perpendicular BD.

Use basic trigonometry to find length of BD.

(This is the height of the triangle.)

Then use Area = 1/2 x base x perpendicular height

So,

A = 1/2 x base x height

A = 1/2 x 12 x 9.511

A = 57.063391

A = 57.1 cm2 (1 dp)

Method 2

Name sides opposite vertices with their lower case letters.

so

A Hollywood film set is being constructed in the desert. The painter is approaching the square based pyramid from behind a sand dune.

The base is 11m, the slant height is 9m and the angle between the base and slant height is 52°.

Calculate the area of the face of the pyramid shown.

What is the length of AC ?

- Name sides opposite vertices with their lower case letters.
- Draw perpendicular CD.
- Use basic trigonometry to find length of CD

This gives

and

So,

Using the letters,

and

Which gives The Sine Rule

A helicopter is approaching an oilrig. The distance from the helicopter to the oilrig is 10 km.

From the bridge of the tanker, the oil rig bears due East and the helicopter bears 040˚ at a distance of 15 km.

What is the bearing of the helicopter from the oil rig ?

Give your answer correct to the nearest degree.

What is the length of BC ?

- Name sides opposite vertices with their lower case letters.

- Try the Sine Rule

No pairs – so can’t use Sine Rule

- Draw perpendicular BD.
- Use basic trigonometry to find lengths of BD and AD

and

- Find length DC

- Use Pythagoras’ Theorem to find BC

This is known as the Cosine Rule

Re-arranging gives

A satellite is being tracked by an observatory telescope and a transmitter.

When the angle of elevation of the telescope is 40˚, the satellite is known to be a distance of 6500 km from the observatory.

The distance between the observatory and the transmitter is 50 km, at a level height.

What is the distance of the satellite from the transmitter?

Give your answer to the nearest kilometer.

Use Cosine rule

Example

The space shuttle is tied down by two 120 m long chains.

The distance between the tie down points on the ground is 100 m.

What is the value of angle θ˚ ? Give your answer correct to 1 decimal place.

AAA – 3 angles

not enough information

AAS – 2 angles, 1 side

Use Sine Rule

ASS – 1 angle ( not included) , 2 sides

Ambiguous case: Either Use Sine Rule twice, or use Cosine Rule with quadratic formula.

Using the Sine Rule:

Using the Cosine Rule

Now solve using the quadratic formula

so

or

Discard the negative value, since the length cannot be negative.