Trigonometry is the study of triangles.

Or from the top angle:

Trigonometric Ratios

The ratios of these sides are given special names:

• sine
• cosine
• tangent

These are shortened to sin, cos and tan.

\[ \begin{alignedat}{2} \sin\theta &= \frac{\text{Opposite}}{\text{Hypotenuse}} \qquad\text{remembered as } S\frac{O}{H}\text{ or SOH} \\[1.4em] \cos\theta &= \frac{\text{Adjacent}}{\text{Hypotenuse}} \qquad\text{remembered as } C\frac{A}{H}\text{ or CAH} \\[1.4em] \tan\theta &= \frac{\text{Opposite}}{\text{Adjacent}} \qquad\text{remembered as } T\frac{O}{A}\text{ or TOA} \end{alignedat} \]

Using Sine to Find a Side

A right-angled triangle is shown. Find x, the length of the side GO.

 

Draw a sketch

The length of side GO is 63.4 m long . (1 dp)

A right-angled triangle is shown. Find x, the length of the side DG.

 

Draw a sketch

The length of side DG is 23.7 m long . (1 dp)

Using Sine to Find an Angle

A right-angled triangle is shown. Find angle θ degrees.

Draw a sketch

\[ \sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \] \[ \sin\theta = \frac{100}{150} \] \[ \text{Make sure your calculator is in Degrees mode} \] \[ \theta = \sin^{-1}\!\left(\frac{100}{150}\right) \] \[ \text{(This is SHIFT–SIN or 2nd–function–SIN on your calculator)} \] \[ \theta = 41.8103^\circ \]

The angle is \(41.8^\circ\) (1dp).

Using Cos to Find a Side

A right-angled triangle is shown. Find x, the length of the side DO.

Draw a sketch

The length of side DO is 129.9 cm long . (1 dp)

Using Cos to Find an Angle

A right-angled triangle is shown. Find angle θ degrees.

Draw a sketch

The angle is \(53.1^\circ\) (1dp).

Using Tan to Find a Side

An aircraft is 800 m directly above a tower on a glide path of \(25^\circ\). How far along the the ground from its touchdown point is the aircraft? Give your answer to 1 d.p.

Draw a sketch

The aircraft is 1715.6 m above the ground.

Using Tan to Find an Angle

An aircraft is 73 m above a building and 200 m from its touchdown point. Calculate the glide‑path angle \(\theta\), correct to 1 decimal place.

The glide‑path angle is \(20.1^\circ\) (1 d.p.).

How high up the wall is the ladder? (Give your answer to 2 d.p.)

Tick what you know: Adjacent
Tick what you want: Opposite

Two ticks → use tangent.

The ladder reaches 1.73 m up the wall (2 d.p.).

R.A Triangle Calculator

Finding the Area of a Triangle

 

Calculate the area of the triangle shown.

\[ \text{Area}_{\triangle} = \tfrac12\, bc \sin A \] \[ \text{Area}_{\triangle} = \tfrac12 \times 9 \times 11 \times \sin 52^\circ \] \[ \text{Area}_{\triangle} = \tfrac12 \times 78.013606 \] \[ \text{Area}_{\triangle} = 39.0065 \] \[ \text{The area of the triangle is } 39\,\text{m}^2 \]

The Sine Rule

There must be one complete top and bottom pair for this to work

Calculate x, the length of side AC.

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \] \[ \frac{12.75}{\sin 97^\circ} = \frac{x}{\sin 58^\circ} = \frac{c}{\sin C} \] \[ 12.75 \sin 58^\circ = x \sin 97^\circ \] \[ x = \frac{12.75 \sin 58^\circ}{\sin 97^\circ} \] \[ x = 10.893 \]

Side AC is 10.89 cm long.

The Cosine Rule

Re-arranging

\[ \begin{alignedat}{2} a^{2} &= b^{2} + c^{2} - 2bc\cos A \\[1em] 2bc\cos A + a^{2} &= b^{2} + c^{2} \\[1em] 2bc\cos A &= b^{2} +c^{2} - a^{2} \\[1em] \cos A &= \frac{b^{2} + c^{2} - a^{2}}{2bc} \end{alignedat} \]

\[ \cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc} \]

 

What is the length of BC?

Try the Sine Rule

No matching pairs → cannot use the Sine Rule.

Using Cosine Rule

\[ \begin{alignedat}{2} a^{2} &= b^{2} + c^{2} - 2bc\cos A \\[1em] a^{2} &= 12^{2} + 10^{2} - 2 \times 12 \times 10 \times \cos 40^\circ \\[1em] &= 144 + 100 - 240\cos 40^\circ \\[1em] &= 244 - 240\cos 40^\circ \\[1em] &= 60.14933365 \\[1.2em] a &= \sqrt{60.14933365} \\[1em] &= 7.755 \\[1em] a &= 7.8\ \text{cm (nearest cm)} \end{alignedat} \]

 

Which Rule to Use