Year 9 Maths Core Trigonometry

Using Trig to Solve Problems

Q: How do you solve trigonometry word problems?

Draw the situation as a right-angled triangle, mark the right angle, the given angle and the known side, label the unknown, then choose sin, cos or tan.

Q: What is an angle of elevation?

It is the angle measured upward from the horizontal to the line of sight to an object above you.

Q: What does a gradient of 1 in 10 mean?

It means the ground rises 1 unit for every 10 units horizontally. The angle with the horizontal is tan inverse of 1 over 10, about 6 degrees.

Q: How do you use trigonometry with an isosceles triangle?

Draw the line from the top to the middle of the base. It makes two identical right-angled triangles, each with half the base and half the top angle.

20 practice questions 2 video lessons Theory + worked examples

This subtopic applies the trigonometric ratios to real situations. In NSW Year 9 Core maths the main work is turning a description into a labelled right-angled triangle and deciding what to find. It brings together the skills of NESA outcome MA5‑TRG‑C‑01.

You will set up triangles for flagpoles and their shadows, ladders, ramps, dam walls and boat anchors, choose the ratio and decide whether the unknown is on the top or the bottom, and find angles as well as sides. You will also convert a gradient such as \(1\) in \(20\) into an angle, handle angles measured from the vertical, use an area to find a missing side first, find exact values like \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), and solve two-part problems with rhombuses and trapeziums. It is part of Year 9 Core Trigonometry.

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Theory

Real problems become right-angled triangles once they are sketched. This Year 9 maths page shows how to turn shadows, ladders, ramps, a gradient such as \(1\) in \(8\), an angle of elevation and isosceles shapes into a labelled triangle, then find a side or an angle with sin, cos or tan.

In a practical problem the right angle is usually where something vertical meets the ground: a wall, a pole, a tree or a mast.

The angle of elevation is the angle measured up from the horizontal to a line of sight, for example from a point on the ground to the top of a building.

A gradient of \(1\) in \(n\) means a rise of \(1\) for every \(n\) measured horizontally, so \(\tan \theta = \dfrac{1}{n}\).

Watch which line an angle is measured from. An angle with the wall and an angle with the ground are the two different acute angles of the same triangle, and they add to \(90^\circ\).

The object is vertical, the shadow lies along the ground and the Sun's rays form the hypotenuse, so \(\tan \theta = \dfrac{\text{height}}{\text{shadow}}\).

The angle of elevation is measured UP from the horizontal to a line of sight. It sits at the observer, with the right angle at the base of the object.

The three ratios, used exactly as before:

\[\sin \theta = \dfrac{\text{opp}}{\text{hyp}} \qquad \cos \theta = \dfrac{\text{adj}}{\text{hyp}} \qquad \tan \theta = \dfrac{\text{opp}}{\text{adj}}\]

\tan θ = \frac{opp}{adj}

A gradient of \(1\) in \(n\):

\[\tan \theta = \dfrac{1}{n}\]

\tan θ = \frac{1}{n}

Find the right angle inside a shape. An isosceles triangle splits into two right-angled triangles down its middle. The diagonals of a rhombus cross at right angles and halve its angles. A rectangle's diagonal makes two right-angled triangles. A cone cut down its middle gives a right-angled triangle with the radius, the height and the slant height. A trapezium splits into a rectangle and a right-angled triangle.

Exact values. For \(45^\circ\), draw a right-angled triangle with both short sides \(1\); its hypotenuse is \(\sqrt{2}\). For \(30^\circ\) and \(60^\circ\), cut an equilateral triangle with sides \(2\) in half: the sides are \(1\), \(\sqrt{3}\) and \(2\). Read any ratio from these triangles as a surd.

How to solve a trigonometry problem

Sketch the situation as a right-angled triangle. Mark the right angle.
Mark the given angle at the vertex it belongs to, and label the known side and the unknown.
Label hyp, opp and adj from the given angle, and choose the ratio.
Solve: multiply or divide for a side, or use an inverse key for an angle.
Answer the question asked, with units, and check the size makes sense. In a two-part problem, carry the unrounded value forward.

Example 1 — A shadow

A tree casts a shadow \(16\) m long when the Sun's rays meet the ground at \(36^\circ\). How tall is the tree, correct to one decimal place?

Solution

The tree is opposite the \(36^\circ\) angle and the shadow is adjacent. Use tan:

\(\tan 36^\circ\)	\(=\)	\(\dfrac{h}{16}\)
\(h\)	\(=\)	\(16 \tan 36^\circ\)
\(\)	\(=\)	\(11.624\ldots\)
\(\)	\(\approx\)	\(\mathbf{11.6} \text{ m}\)

h = 16 \tan 36 ° \approx 11.6

Example 2 — An angle with the wall

A \(5.6\) m ladder leans against a wall, making \(25^\circ\) with the WALL. How far up the wall does it reach, correct to one decimal place?

Solution

The angle is at the top, between the ladder and the wall. From THAT angle the wall is the adjacent side and the ladder is the hypotenuse:

\(\cos 25^\circ\)	\(=\)	\(\dfrac{w}{5.6}\)
\(w\)	\(=\)	\(5.6 \cos 25^\circ\)
\(\)	\(=\)	\(5.075\ldots\)
\(\)	\(\approx\)	\(\mathbf{5.1} \text{ m}\)

w = 5.6 \cos 25 ° \approx 5.1

\(5.1 < 5.6\): the ladder reaches less high than its own length, as it must.

Example 3 — A gradient of 1 in 8

A path has a gradient of \(1\) in \(8\). What angle does it make with the horizontal, correct to the nearest degree?

Solution

\(1\) in \(8\) means a rise of \(1\) for every \(8\) along the horizontal. Those are the opposite and adjacent sides:

\(\tan \theta\)	\(=\)	\(\dfrac{1}{8}\)
\(\theta\)	\(=\)	\(\tan^{-1}\left(\dfrac{1}{8}\right)\)
\(\)	\(=\)	\(7.125\ldots^\circ\)
\(\)	\(\approx\)	\(\mathbf{7^\circ}\)

θ = \tan^{- 1} (\frac{1}{8}) \approx 7 °

Example 4 — Split a shape into right-angled triangles

The front of a tent is an isosceles triangle. Its two sloping sides are \(2.6\) m long and each makes \(58^\circ\) with the ground. Find the width of the base, correct to two decimal places.

Solution

A line from the top to the middle of the base cuts the triangle into two identical right-angled triangles. In one of them, \(2.6\) is the hypotenuse and half the base is adjacent to \(58^\circ\):

\(\cos 58^\circ\)	\(=\)	\(\dfrac{\text{half}}{2.6}\)
\(\text{half}\)	\(=\)	\(2.6 \cos 58^\circ = 1.3777\ldots\)
\(\text{width}\)	\(=\)	\(2 \times 1.3777\ldots\)
\(\)	\(\approx\)	\(\mathbf{2.76} \text{ m}\)

width = 2 \times 2.6 \cos 58 ° \approx 2.76

Double the unrounded half, then round.

Common pitfalls

Read which angle is given. \(20^\circ\) to the wall is not \(20^\circ\) to the ground. A quick sketch settles it.

The hypotenuse is the longest side. A ladder cannot reach higher up a wall than its own length. An answer like that means the wrong ratio was used.

Halve before you use the ratio. In an isosceles triangle, rhombus or cone, the right-angled triangle uses HALF the base and HALF the angle at the top.

Answer what was asked. Some questions want a height, some a distance along the ground, and some the angle itself.

Frequently asked questions

How do you solve trigonometry word problems?

Draw the situation as a right-angled triangle, mark the right angle, the given angle and the known side, label the unknown, then choose sin, cos or tan.

What is an angle of elevation?

It is the angle measured upward from the horizontal to the line of sight to an object above you.

What does a gradient of 1 in 10 mean?

It means the ground rises \(1\) unit for every \(10\) units horizontally. The angle with the horizontal is \(\tan^{-1}\left(\dfrac{1}{10}\right) \approx 6^\circ\).

How do you use trigonometry with an isosceles triangle?

Draw the line from the top to the middle of the base. It makes two identical right-angled triangles, each with half the base and half the top angle.

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Using Trig to Solve Problems

📖 Theory

How to solve a trigonometry problem

Common pitfalls

Frequently asked questions

Theory