2D Applications
2D Applications in NSW Year 10 Core maths uses right-angled trigonometry for real problems, many of them using angles of elevation and depression (NESA outcomes MA5‑TRG‑C‑01 and MA5‑TRG‑C‑02). An angle of elevation is measured up from the horizontal and an angle of depression down from it, and for the same line of sight the two are equal.
You will find the heights of towers, cliffs and trees, distances to boats and runways, and the angle of the sun from a shadow. You will also solve problems that need two right-angled triangles, such as two boats seen from a lighthouse, two buildings of different heights and a marker moved closer to a building, and find the angle at the centre of a circle from a chord. It is part of Year 10 Core Trigonometry.
Every question with a fully worked solution.
- 2D Applications - Video - Angles of Elevation and Depression Watch
Theory
Trigonometry problems in real settings start with a sketch: find the right angle, mark the given angle, then use sine, cosine or tangent. This Year 10 maths page covers angles of elevation and depression, finding an angle with the inverse keys, eye height, isosceles shapes and problems that need two right-angled triangles.
An angle of elevation is the angle you look up through, measured from the horizontal. An angle of depression is the angle you look down through, measured from the horizontal.
Both are measured from a horizontal line, never from a vertical wall. Because the horizontal at the top is parallel to the ground, the angle of depression from \(A\) to \(B\) equals the angle of elevation from \(B\) to \(A\) (alternate angles).
To find an angle, write the ratio of two known sides, then use the inverse key: \(\theta = \sin^{-1}(\ldots)\), \(\cos^{-1}(\ldots)\) or \(\tan^{-1}(\ldots)\).
Many shapes hide a right-angled triangle. A height drawn in an isosceles triangle halves the base and the apex angle. A diagonal splits a rectangle into two right-angled triangles.
In any right-angled triangle, relative to the angle \(\theta\):
To find an angle from two sides:
How to solve a practical trigonometry problem
- Sketch the situation. Walls, poles and trees meet level ground at \(90^\circ\). Draw the horizontal for any angle of elevation or depression.
- Find the right-angled triangle, and move an angle of depression to the bottom of the triangle with alternate angles.
- Label the sides from the angle and choose sine, cosine or tangent (or an inverse to find an angle).
- Solve, keep full calculator values, and add or subtract any extra lengths at the end.
Common set-ups
- Shadows: the object is opposite the sun's angle of elevation and the shadow is adjacent. With both lengths known, \(\theta = \tan^{-1}\left(\dfrac{\text{height}}{\text{shadow}}\right)\).
- Eye height: the triangle starts at eye level. Add the eye height to the answer.
- From the top of a building: the horizontal from the roof meets the other building above the ground. Add the roof height (Example 3).
- Something moves closer (a marker, a boat, an observer): find the distance in each position, then subtract.
- Two objects on opposite sides of a tower or lighthouse: find each distance, then add. On the same side, subtract.
- A snapped tree or post: the upright part is opposite the angle and the broken part is the hypotenuse. The original height is the two parts added.
- A path in two stages (two slopes): solve each stage, then add the heights and add the horizontal distances.
- Chord of a circle: the radii to the chord's ends make an isosceles triangle. The perpendicular from the centre halves the chord and the angle at the centre, so find the half-angle and double it.
The horizontal at the top is parallel to the ground, so the angle at the car is also \(24^\circ\) (alternate angles). From that angle, the tower is opposite and \(d\) is adjacent.
| \(\tan 24^\circ\) | \(=\) | \(\dfrac{28}{d}\) |
| \(d\) | \(=\) | \(\dfrac{28}{\tan 24^\circ}\) |
| \(\) | \(=\) | \(62.8890\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{62.9}\text{ m}\) |
Never put the angle of depression inside the triangle at the top. It sits between the line of sight and the horizontal, not the tower.
(i) the ground
(ii) the wall.
The ladder is the hypotenuse. From the angle at the ground, the height up the wall is opposite, so use sine and then the inverse key.
(i)
| \(\sin\theta\) | \(=\) | \(\dfrac{4.5}{4.8}\) |
| \(\theta\) | \(=\) | \(\sin^{-1}\left(\dfrac{4.5}{4.8}\right)\) |
| \(\) | \(=\) | \(69.63\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{70^\circ}\) |
(ii) The three angles of the triangle add to \(180^\circ\), so the two acute angles add to \(90^\circ\).
| \(\alpha\) | \(=\) | \(90^\circ - 69.63\ldots^\circ\) |
| \(\) | \(=\) | \(20.36\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{20^\circ}\) |
Read the question carefully: an angle with the wall is not the same as an angle with the ground.
(i) the distance \(d\) between the building and the tower
(ii) the height \(H\) of the tower.
The horizontal from the roof splits the picture into two right-angled triangles. The lower one gives \(d\). The upper one uses \(d\) to find the part of the tower above roof level.
(i) Lower triangle: \(12\) m is opposite \(19^\circ\) and \(d\) is adjacent.
| \(d\) | \(=\) | \(\dfrac{12}{\tan 19^\circ}\) |
| \(\) | \(=\) | \(34.8505\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{34.9}\text{ m}\) |
(ii) Upper triangle: the part above the roof is opposite \(34^\circ\).
| \(x\) | \(=\) | \(34.8505\ldots \times \tan 34^\circ\) |
| \(\) | \(=\) | \(23.5069\ldots\) |
| \(H\) | \(=\) | \(23.5069\ldots + 12\) |
| \(\) | \(=\) | \(35.5069\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{35.5}\text{ m}\) |
Add the \(12\) m below roof level: the upper triangle only reaches down to the horizontal line.
(i) the height \(h\) of the truss
(ii) the length \(r\) of each rafter.
The height from the apex meets the base at right angles and halves it. Each half is a right-angled triangle with base \(4.6\) m.
(i) \(h\) is opposite \(24^\circ\) and \(4.6\) is adjacent.
| \(h\) | \(=\) | \(4.6 \tan 24^\circ\) |
| \(\) | \(=\) | \(2.0480\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{2.05}\text{ m}\) |
(ii) \(r\) is the hypotenuse and \(4.6\) is adjacent.
| \(\cos 24^\circ\) | \(=\) | \(\dfrac{4.6}{r}\) |
| \(r\) | \(=\) | \(\dfrac{4.6}{\cos 24^\circ}\) |
| \(\) | \(=\) | \(5.0353\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{5.04}\text{ m}\) |
Use half the span, not the whole \(9.2\) m. If a question asks for a whole base, double the half you found.
Common pitfalls
Frequently asked questions
What is the difference between angle of elevation and angle of depression?
An angle of elevation is measured up from the horizontal to an object above you. An angle of depression is measured down from the horizontal to an object below you. From two points looking at each other, the two angles are equal.
Why is the angle of depression equal to the angle of elevation?
The horizontal line at the top is parallel to the level ground. The line of sight crosses both parallel lines, so the angle of depression and the angle at the bottom are alternate angles, which are equal.
How do you find an angle in a right-angled triangle?
Write the ratio of the two known sides using sine, cosine or tangent, then use the inverse key, for example \(\theta = \tan^{-1}\left(\dfrac{7}{10}\right)\). Round to the degree or minute asked for.
How do you find the angle of elevation of the sun?
The object and its shadow form a right angle. The height is opposite and the shadow is adjacent, so \(\theta = \tan^{-1}\left(\dfrac{\text{height}}{\text{shadow}}\right)\).
How do you find the height of a building using trigonometry?
Measure your distance from the base and the angle of elevation of the top. The height above eye level is distance \(\times \tan\theta\). Add your eye height to get the full height.