What does a gradient of 1 to 12 mean?

It means 1 unit of vertical rise for every 12 units of horizontal run. The angle of inclination is theta equals tan inverse of one twelfth, which is approximately 4.76 degrees. This is a common road-sign notation in Australia.

What does a gradient of 1 to 12 mean?

It means 1 unit of vertical rise for every 12 units of horizontal run. The angle of inclination is theta equals tan inverse of one twelfth, which is approximately 4.76 degrees. This is a common road-sign notation in Australia.

Year 11 General Applications Of Trigonometry

Applications Of Right-Angled Triangles To Problem-Solving

Q: How do I find the angle a body diagonal makes with the base?

Two steps. Step 1: use Pythagoras on the base rectangle to find the base diagonal length b equals the square root of length squared plus width squared. Step 2: the body diagonal, the base diagonal and the vertical height form a 2D right triangle. The angle with the base is theta equals tan inverse of height divided by b.

Q: What if the angle is measured from the vertical instead of the horizontal?

Two options. Either subtract from 90 degrees first to get the angle from the horizontal, then use SOH-CAH-TOA as normal. Or use the angle as given, but swap which side is opposite — the side opposite the vertical-angle is the horizontal side, not the vertical one.

11 practice questions 2 video lessons Theory + worked examples

Theory

Real-world right-triangle problems all reduce to four steps: sketch the situation, label the sides, choose the right trig ratio, then solve. Common contexts include ladders, ramps, shadows, surveying and 3D body diagonals.

Right-triangle trigonometry turns up whenever a real-world situation hides a right angle. The challenge is to spot the right triangle, label its sides correctly, and pick the trig ratio that connects what you know to what you want.

Some common situations and where the right angle sits:

Ladder against a wall — right angle between wall and ground.
Tree and shadow — right angle between tree and ground.
Ramp or incline — right angle between vertical rise and horizontal run.
Surveying across a river — right angle where the line of sight meets the bank.
3D body diagonal — right angle between a vertical edge and the base diagonal.

A gradient like $1 : 12$ means $1$ m of vertical rise per $12$ m of horizontal run. The angle of inclination is $θ = \tan^{- 1} (1 / 12)$ .

Ladder problem: right angle at the base of the wall

Incline:

\tan θ = \frac{rise}{run}

The trig ratios in real-world problems

Same three ratios as always — what changes is which physical thing maps to opposite, adjacent and hypotenuse:

\sin θ = \frac{Opp}{Hyp} \cos θ = \frac{Adj}{Hyp} \tan θ = \frac{Opp}{Adj}

\sin θ = \frac{Opp}{Hyp}, \cos θ = \frac{Adj}{Hyp}, \tan θ = \frac{Opp}{Adj}

Gradient as an angle

A road gradient written as a ratio $a : b$ means $a$ units of rise for every $b$ units of run:

θ = \tan^{- 1} (\frac{a}{b})

θ = \tan^{- 1} (\frac{a}{b})

3D body diagonal

To find the angle a body diagonal makes with the base of a rectangular box, first find the base diagonal with Pythagoras, then treat it as one side of a 2D right triangle with the box's height:

b = \sqrt{l^{2} + w^{2}} \tan θ = \frac{height}{b}

b = \sqrt{l^{2} + w^{2}}, \tan θ = \frac{height}{b}

The four-step approach for any application

Sketch the situation and mark the right angle clearly.
Label the sides relative to the known (or unknown) angle: Opp, Adj, Hyp.
Choose the trig ratio that uses the two sides you care about — one known, one unknown (or two known sides if finding an angle).
Substitute and solve, then sanity-check the answer. (For example, a ladder height should be less than the ladder length.)

Example 1 — Ladder problem (find side)

A ladder leans against a wall. Its base is

9.5

m from the wall, and it makes a

60^{\circ}

angle with the ground. Find the length of the ladder

L

(1 dp).

Solution

Sketch the situation:

The ladder is the hypotenuse, $9.5$ m is adjacent to $60^{\circ}$ . Use $\cos$ :

$\cos 60^{\circ}$	$=$	$\frac{9.5}{L}$
$L$	$=$	$\frac{9.5}{\cos 60^{\circ}} = \frac{9.5}{0.5}$
$L$	$=$	$19.0 m$

L = \frac{9.5}{\cos 60 °} = 19.0 m

Example 2 — Find an angle (ladder)

5

m ladder reaches

4.2

m up a wall. Find the angle

θ

between the ladder and the wall (nearest minute).

Solution

Sketch — $θ$ is at the top, between the ladder and the wall:

From $θ$ at the top: $4.2$ m (down the wall) is adjacent, $5$ m (the ladder) is the hypotenuse. Use $\cos$ :

$\cos θ$	$=$	$\frac{4.2}{5} = 0.84$
$θ$	$=$	$\cos^{- 1} (0.84)$
$θ$	$\approx$	$32^{\circ} 52^{'}$

θ = \cos^{- 1} (0.84) \approx 32 ° 52'

Example 3 — Train on incline

A train climbs an incline of

4^{\circ}

, travelling

4

km horizontally. How much altitude has it gained (2 dp)?

Solution

Sketch the right triangle:

Altitude is opposite to $4^{\circ}$ , horizontal is adjacent. Use $\tan$ :

$\tan 4^{\circ}$	$=$	$\frac{h}{4}$
$h$	$=$	$4 \tan 4^{\circ}$
$h$	$\approx$	$0.28 km$

h = 4 \tan 4 ° \approx 0.28 km

Example 4 — 3D body diagonal angle

A box is

6 \times 4 \times 3

cm. Find the angle between the body diagonal and the base (nearest degree).

Solution

Two-step problem. Sketch the box with the body diagonal and base diagonal marked:

Step 1 — base diagonal using Pythagoras:

$b^{2}$	$=$	$6^{2} + 4^{2} = 52$
$b$	$=$	$\sqrt{52} \approx 7.21$

Step 2 — angle with height $3$ opposite and $b$ adjacent:

$\tan θ$	$=$	$\frac{3}{7.21}$
$θ$	$\approx$	$23^{\circ}$

b = \sqrt{52}, \tan θ = \frac{3}{b}, θ \approx 23 °

Common pitfalls

Read the angle's location carefully. If a problem says "the ladder makes an angle of

17^{\circ}

with the pole", the angle sits at the top (between ladder and vertical wall), not at the bottom. Sketch first.

Angle from the vertical, not the horizontal. Some problems measure the angle from the vertical (e.g.\ an anchor rope at

23^{\circ}

from vertical). Either add/subtract from

90^{\circ}

to get the angle from horizontal, or just use the one you're given directly with the correct sides.

3D problems need a two-step approach. For a body diagonal of a box, find the base diagonal first with Pythagoras, then treat that as one side of a 2D right triangle with the height. (Same trick as 3D Pythagoras.)

Sanity-check the answer. A ladder height should be less than the ladder length. A train's altitude gain over a few km on a gentle slope should be small. If your answer looks impossible, recheck which sides are opposite/adjacent/hypotenuse.

Frequently asked questions

How do I solve a ladder problem?

Sketch the right triangle: the wall is vertical, the ground is horizontal, the ladder is the hypotenuse. Label the angle the ladder makes with the ground. Then pick the trig ratio that uses the side you know and the side you want — $\sin$ for height up the wall and ladder length, $\cos$ for distance from wall and ladder length, $\tan$ for height and distance.

What does a gradient of

1 : 12

mean?

It means $1$ unit of vertical rise for every $12$ units of horizontal run. The angle of inclination is $θ = \tan^{- 1} (1 / 12) \approx {4.76}^{\circ}$ . This is a common road-sign notation in Australia.

How do I find the angle a body diagonal makes with the base?

Two steps. Step 1: use Pythagoras on the base rectangle to find the base diagonal length $b = \sqrt{l^{2} + w^{2}}$ . Step 2: the body diagonal, the base diagonal and the vertical height form a 2D right triangle. The angle with the base is $θ = \tan^{- 1} (height / b)$ .

What if the angle is measured from the vertical instead of the horizontal?

Two options. Either subtract from $90^{\circ}$ first to get the angle from the horizontal, then use SOH-CAH-TOA as normal. Or use the angle as given, but swap which side is "opposite" — the side opposite the vertical-angle is the horizontal side, not the vertical one.

Why does my answer for a ladder height seem larger than the ladder?

That's a sign you've used the wrong trig ratio or flipped opposite/adjacent. The vertical height a ladder reaches must always be less than the ladder length (it's only equal if the ladder is exactly vertical). Re-sketch, recheck which side is the hypotenuse, and try again.

Do I sketch every problem?

Yes — always. A quick rough sketch with the right angle marked and the angle/sides labelled is the most reliable way to avoid using the wrong ratio. Most marks lost in trig word problems come from skipping the sketch step.