Right-Angled Trigonometry

In short: Right-angled trigonometry links the angles and sides of a right-angled triangle with SOH CAH TOA: $sin θ = \frac{opposite}{hypotenuse}$ , $cos θ = \frac{adjacent}{hypotenuse}$ , $tan θ = \frac{opposite}{adjacent}$ . Inverse trig finds an angle from two sides.

Starting point: Years 9 and 10

Pythagoras' theorem

In a right-angled triangle with hypotenuse $c$ (the longest side, opposite the right angle):

$a^{2} + b^{2} = c^{2}$

So a triangle with sides 6 and 8 has hypotenuse $36 + 64 = 10$ .

Naming the sides

Pick an angle $θ$ . Then:

Hypotenuse (H): the longest side, opposite the right angle.
Opposite (O): across from $θ$ .
Adjacent (A): next to $θ$ , not the hypotenuse.

The opposite and adjacent sides depend on which angle you pick.

The three ratios

One ratio gives the others. If $tan θ = \frac{3}{4}$ and $θ$ is acute, draw a right triangle with opposite 3 and adjacent 4. Pythagoras gives the hypotenuse 5, so $sin θ = \frac{3}{5}$ and $cos θ = \frac{4}{5}$ .

Angles of elevation and depression

The angle of elevation is measured up from the horizontal. The angle of depression is measured down from the horizontal. They are equal for two people looking at each other, because they are alternate angles between parallel lines.

Both angles are measured from a horizontal line, never from the cliff. The horizontal at the top and the sea are parallel, so the two angles are equal alternate angles.

Eye height. An angle of elevation measured from someone's eyes gives the height above their eyes. Add their eye height at the end: if the triangle gives 26.2 m and the eyes are 1.6 m above the ground, the tree is 27.8 m tall.

Trigonometry gives the height above the eyes (26.2 m). The eyes are 1.6 m above the ground, so add that at the end: 27.8 m.

Bearings

A true bearing is a three-digit angle measured clockwise from north. East is $09 0^{\circ}$ , south is $18 0^{\circ}$ , west is $27 0^{\circ}$ . A compass bearing like S $4 0^{\circ}$ W means start facing south and turn $4 0^{\circ}$ toward west.

To change a compass bearing to a true bearing, start from the true bearing of the first direction. S $4 0^{\circ}$ W is $18 0^{\circ} + 4 0^{\circ} = 22 0^{\circ}$ . S $5 0^{\circ}$ E is $18 0^{\circ} - 5 0^{\circ} = 13 0^{\circ}$ . N $3 0^{\circ}$ W is $36 0^{\circ} - 3 0^{\circ} = 33 0^{\circ}$ .

Right-angled trig

Pythagoras

a² + b² = c²

Sides

Hypotenuse
Opposite
Adjacent

Ratios

sin = O/H
cos = A/H
tan = O/A

Uses

Elevation and depression
Bearings

Exact values

45° triangle
30° and 60° triangle

Year 11Exact values

Two special triangles give exact ratios with no calculator.

Half a square: sides 1, 1, $2$ with angles $4 5^{\circ}$ , $4 5^{\circ}$ , $9 0^{\circ}$ .

Half an equilateral triangle: sides 1, $3$ , 2 with angles $3 0^{\circ}$ , $6 0^{\circ}$ , $9 0^{\circ}$ .

Cut a square of side 1 along its diagonal, or an equilateral triangle of side 2 down its middle. Pythagoras gives the missing sides: √(1² + 1²) = √2 and √(2² − 1²) = √3.

	$3 0^{\circ}$	$4 5^{\circ}$	$6 0^{\circ}$
sin	$\frac{1}{2}$	$\frac{1}{2}$	$\frac{3}{2}$
cos	$\frac{3}{2}$	$\frac{1}{2}$	$\frac{1}{2}$
tan	$\frac{1}{3}$	1	$3$

Notice that $sin 3 0^{\circ} = cos 6 0^{\circ}$ . In general $sin θ = cos (9 0^{\circ} - θ)$ . These are complementary angles.

Year 11Problems in three dimensions

In 3D questions, find a right-angled triangle hiding in the shape and solve it, then use that answer in a second triangle.

Example. A box is 4 by 3 by 12. The floor diagonal is $4^{2} + 3^{2} = 5$ . The space diagonal is $5^{2} + 1 2^{2} = 13$ . The angle it makes with the floor is $tan^{- 1} (\frac{12}{5}) \approx 67. 4^{\circ}$ .

First the floor triangle: 4, 3 and the floor diagonal 5. Then the upright triangle: 5 along the floor, 12 up the edge, and the space diagonal 13. The angle θ is in the upright triangle.

Two sightings. From $A$ the top of a tower has elevation $3 0^{\circ}$ ; walking 40 m closer to $B$ , it is $6 0^{\circ}$ . With height $h$ : from $B$ , the distance to the tower is $\frac{h}{tan 6 0 ^{\circ}} = \frac{h}{3}$ , and from $A$ it is $\frac{h}{tan 3 0 ^{\circ}} = 3 h$ . The difference is 40: $3 h - \frac{h}{3} = \frac{2 h}{3} = 40$ , so $h = 203$ m exactly.

Two right-angled triangles share the height h. The distance from A is √3h and from B it is h ÷ √3. Their difference is the 40 m walked.

Bearings in 3D. If $C$ is due south of the foot of a tower and $D$ is due west, then angle $C B D$ at the foot is $9 0^{\circ}$ . Find $B C$ and $B D$ from the elevations, then $C D$ by Pythagoras on the ground. A point 30 m south and 40 m east of the foot is $3 0^{2} + 4 0^{2} = 50$ m away.

Each line of sight makes an upright right-angled triangle with the tower: use α to find BD and β to find BC. West and south are at right angles, so the shaded ground triangle gives CD by Pythagoras.

Quick summary

Pythagoras: $a^{2} + b^{2} = c^{2}$ , with $c$ the hypotenuse.
SOH CAH TOA links an angle to two sides.
Use inverse trig ( $sin^{- 1}$ , $cos^{- 1}$ , $tan^{- 1}$ ) to find angles.
Elevation is up from the horizontal, depression is down. True bearings go clockwise from north.
Exact values come from the $4 5^{\circ}$ and $3 0^{\circ}$ - $6 0^{\circ}$ triangles.

Quiz

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Try 3 questions now

A right-angled triangle has shorter sides 5 and 12. The hypotenuse is:
1. A 17
2. B 13
3. C 7
4. D 169
Show answer
B: 13
$25 + 144 = 169 = 13$ .
Which ratio is $\frac{opposite}{hypotenuse}$ ?
1. A $sin θ$
2. B $cos θ$
3. C $tan θ$
4. D $θ$
Show answer
A: $sin θ$
SOH.
A ladder 5 m long leans at $6 0^{\circ}$ to the ground. How high up the wall does it reach?
1. A 2.5 m
2. B 4.33 m
3. C 5.77 m
4. D 8.66 m
Show answer
B: 4.33 m
$h = 5 sin 6 0^{\circ} \approx 4.33$ m.