Applying Trigonometry
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 \(8\), angles measured from a wall, 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.
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 three ratios, used exactly as before:
A gradient of \(1\) in \(n\):
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.
The tree is opposite the \(36^\circ\) angle and the shadow is adjacent. Use tan:
| \(\tan 36^\circ\) | \(=\) | \(\dfrac{h}{16}\) |
| \(h\) | \(=\) | \(16 \tan 36^\circ\) |
| \(\) | \(=\) | \(11.624\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{11.6} \text{ m}\) |
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:
| \(\cos 25^\circ\) | \(=\) | \(\dfrac{w}{5.6}\) |
| \(w\) | \(=\) | \(5.6 \cos 25^\circ\) |
| \(\) | \(=\) | \(5.075\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{5.1} \text{ m}\) |
\(5.1 < 5.6\): the ladder reaches less high than its own length, as it must.
\(1\) in \(8\) means a rise of \(1\) for every \(8\) along the horizontal. Those are the opposite and adjacent sides:
| \(\tan \theta\) | \(=\) | \(\dfrac{1}{8}\) |
| \(\theta\) | \(=\) | \(\tan^{-1}\left(\dfrac{1}{8}\right)\) |
| \(\) | \(=\) | \(7.125\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{7^\circ}\) |
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 \(58^\circ\):
| \(\cos 58^\circ\) | \(=\) | \(\dfrac{\text{half}}{2.6}\) |
| \(\text{half}\) | \(=\) | \(2.6 \cos 58^\circ\) |
| \(\) | \(=\) | \(1.3777\ldots\) |
| \(\text{width}\) | \(=\) | \(2 \times 1.3777\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{2.76} \text{ m}\) |
Double the unrounded half, then round.
Common pitfalls
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 if the angle is measured from the wall or the vertical?
Mark it at the top, between the vertical line and the sloping side. From that angle the vertical side is the adjacent side. Or subtract it from \(90^\circ\) to get the angle with the ground.
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.