Angles of Elevation & Depression - Extended Maths

Defining each angle

Both angles are measured from a horizontal line, the flat line of sight straight ahead. The only difference is whether you tilt your eyes up or down.

Angle of elevation: The angle between the horizontal and your line of sight when you look up at a higher object.

Angle of depression: The angle between the horizontal and your line of sight when you look down at a lower object.

Elevation and depression are equal

Looking from A up to B, and from B down to A, gives the same number. The two horizontals are parallel, so the angle of elevation and the angle of depression are alternate angles, hence equal. This lets you move the angle into whichever triangle is easier to work with.

Drawing the diagram

Nearly every question is solved the moment you have a good sketch. Turn the words into a right-angled triangle in three steps:

Draw the ground as a horizontal line and the object as a vertical line, meeting at a right angle.
Mark the observer, then draw the sloped line of sight to make the hypotenuse.
Label the angle at the correct spot: elevation sits at the observer looking up; depression sits at the top, measured down from a horizontal drawn there.

Now decide which sides you know and which you want, and pick sine, cosine or tangent with SOHCAHTOA. In most problems the horizontal distance and the height are the two sides involved, so tangent is the usual choice.

Solving elevation

Worked example

Standing 20 m from the base of a tree, the angle of elevation to the top is \(35^\circ\). How tall is the tree? Give your answer to 1 decimal place.

Sketch: the 20 m is the horizontal (adjacent), the tree height is the vertical (opposite), and \(35^\circ\) sits at the observer.

Opposite and adjacent means tangent: \(\tan 35^\circ = \dfrac{h}{20}\).

Rearrange: \(h = 20 \times \tan 35^\circ = 20 \times 0.7002\ldots = 14.00\ldots\)

Height \(\approx 14.0\) m

Solving depression

Worked example

From the top of a 40 m cliff, the angle of depression to a boat at sea is \(25^\circ\). How far is the boat from the base of the cliff? Give your answer to 1 decimal place.

Sketch: the cliff height (40 m) is vertical, the sea distance is horizontal. The \(25^\circ\) depression at the top equals the \(25^\circ\) angle of elevation at the boat (alternate angles).

Working at the boat, the 40 m is opposite the angle and the distance \(x\) is adjacent, so use tangent: \(\tan 25^\circ = \dfrac{40}{x}\).

Rearrange: \(x = \dfrac{40}{\tan 25^\circ} = \dfrac{40}{0.4663\ldots} = 85.78\ldots\)

Distance \(\approx 85.8\) m

Where this is assessed

These are textbook Criterion D (Applying maths in real-life contexts) problems, since they wrap trigonometry in a real scene. A clear labelled diagram and each step shown pick up Criterion C (Communicating), while the trig calculation itself is Criterion A (Knowing and understanding). Always draw the diagram before reaching for the calculator.

Check yourself

1. What is the difference between an angle of elevation and an angle of depression? +

Both are measured from the horizontal. An angle of elevation is when you look up at a higher object; an angle of depression is when you look down at a lower one. Between the same two points they are equal, being alternate angles between parallel horizontals.

2. From 50 m away, the angle of elevation to the top of a mast is \(30^\circ\). Find the height (to 1 d.p.). +

Tangent links the height (opposite) and the 50 m (adjacent): \(\tan 30^\circ = \dfrac{h}{50}\), so \(h = 50 \times \tan 30^\circ = 50 \times 0.5774 = 28.86\ldots\) The height is \(\mathbf{28.9}\) m.

3. From a 60 m tower, the angle of depression to a car is \(40^\circ\). How far is the car from the base (to 1 d.p.)? +

The depression equals the elevation at the car, where 60 m is opposite and the distance \(x\) is adjacent: \(\tan 40^\circ = \dfrac{60}{x}\), so \(x = \dfrac{60}{\tan 40^\circ} = \dfrac{60}{0.8391} = 71.50\ldots\) The distance is \(\mathbf{71.5}\) m.

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