IB Applications and Interpretation SL · Trigonometry · Section 2 of 4

Trigonometric ratios in the right-angled triangle

3 worked examples · By Álvaro López

Key idea

Sine, cosine and tangent relate an acute angle of a right-angled triangle to the ratio of two of its sides. With them you can calculate a distance you cannot measure (the height of a building, the length of a ladder) from one known angle and one known side. In the IB exam they are the starting tool for almost every problem with a diagram.

The three sides, depending on the angle

In a right-angled triangle, the longest side, the one opposite the right angle, is the hypotenuse. The other two are the legs. Their names, "opposite" and "adjacent", depend on the acute angle $θ$ you choose as reference:

the opposite side is the one across from $θ$ , not touching it;
the adjacent side is the one that forms $θ$ together with the hypotenuse.

Right-angled triangle showing the hypotenuse, the opposite side and the adjacent side to angle theta

If you change the angle, the two legs swap roles; the hypotenuse never changes.

These ratios depend only on the angle, not on the size of the triangle: if you enlarge the triangle, all the sides are multiplied by the same factor and the ratio does not change. That is why a calculator can store the value of $sin 3 0^{\circ}$ and be used for any triangle with that angle.

To remember which sides go in each ratio, use the rule SOH-CAH-TOA: Sine = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, Tangent = Opposite / Adjacent. As a memory aid: "Some Old Horses Caught Another Horse Taking Oats Away".

Finding a side and finding an angle

Finding a side. You know one acute angle and one side. Choose the ratio that contains that side and the side you are looking for, and rearrange. If the side you are looking for ends up in the denominator, the equation is solved in the same way: cross-multiply.

Finding an angle. You know two sides. Calculate their ratio and apply the inverse function. If $sin θ = 0.6$ , then $θ = sin^{- 1} (0.6) \approx 36. 9^{\circ}$ . The functions $sin^{- 1}$ , $cos^{- 1}$ and $tan^{- 1}$ return the acute angle that has that ratio.

Solving a right-angled triangle

Solving a triangle means calculating all its sides and angles. In a right-angled triangle you only need two pieces of information besides the right angle, and one of them must be a side:

If you know two sides, the third comes from Pythagoras, one acute angle from an inverse ratio, and the other as $9 0^{\circ}$ minus the first.
If you know one side and one acute angle, the other angle is $9 0^{\circ}$ minus the given one, and the sides come from the ratios.

When a triangle is not right-angled, it can sometimes be split into two with a height. This is the idea behind composite figures and 3D problems.

Angles of elevation and depression

Many questions give angles measured "looking up" or "looking down". They are always measured from the horizontal through the observer's eye, never from the vertical.

An observer with their horizontal, the angle of elevation to a higher point and the angle of depression to a lower point

The horizontals through the observer and through the object are parallel. So, if from the top of a tower you see a car at a depression of $2 0^{\circ}$ , from the car you see the top of the tower at an elevation of $2 0^{\circ}$ (alternate interior angles between parallel lines). This fact lets you place the same angle inside the right-angled triangle you draw.

Problems with diagrams

Most IB problems describe a situation and do not mention any triangle. Your job is to find the hidden right-angled triangle.

Ladder leaning against a wall: the wall (vertical) and the ground (horizontal) form the right angle; the ladder is the hypotenuse.
Pitched roof: the height of the ridge divides the triangular gable into two identical right-angled triangles.
Shadows, towers and trees: the height, the horizontal distance and the line of sight form the triangle.
Bearings and distances: moving north and then east forms a right angle; when the question says "N $3 0^{\circ}$ E", it means you start from north and turn $3 0^{\circ}$ towards the east.

Simple 3D figures. A prism or a box contains many right-angled triangles. The technique is to draw only the flat triangle you need and calculate step by step. For example, in a box with a base of $8 \times 6$ and height $5$ (in cm), the diagonal of the base is $8^{2} + 6^{2} = 10$ cm. Then the diagonal of the box is the hypotenuse of a triangle with legs $10$ and $5$ , that is $125 \approx 11.2$ cm, and this diagonal makes an angle $tan^{- 1} \frac{5}{10} \approx 26. 6^{\circ}$ with the base.

Worked examples

Example 1 Basic

Solving a right-angled triangle

Triangle $A B C$ is right-angled at $B$ , with $A B = 7.5$ cm and $A C = 12$ cm.

a) Calculate angle $A$ in degrees.

b) Calculate the length $B C$ .

c) Calculate angle $C$ .

Solution

The hypotenuse is $A C = 12$ (opposite the right angle).

a) With respect to angle $A$ , side $A B$ is adjacent and $A C$ is the hypotenuse, so we use cosine: $cos A = \frac{7.5}{12} \Rightarrow A = cos^{- 1} (\frac{7.5}{12}) \approx 51. 3^{\circ}$

b) By Pythagoras, $B C^{2} = 1 2^{2} - 7. 5^{2} = 144 - 56.25 = 87.75$ , so $B C = 87.75 \approx 9.37$ cm.

c) The acute angles add up to $9 0^{\circ}$ : $C = 9 0^{\circ} - 51. 3^{\circ} = 38. 7^{\circ}$ .

Check: $sin A = \frac{9.37}{12} \approx 0.781$ and $sin 51. 3^{\circ} \approx 0.781$ .

Example 2 Medium

A leaning ladder

A ladder 5.2 m long leans against a vertical wall. It makes an angle of $6 8^{\circ}$ with the horizontal ground.

a) Calculate the height on the wall that the ladder reaches.

b) Calculate the distance between the foot of the ladder and the wall.

c) To reach a window 4.5 m high with the same ladder, what angle must it make with the ground?

Solution

The ladder is the hypotenuse (5.2 m); the height on the wall is the side opposite the angle with the ground, and the distance to the foot is the adjacent side.

a) $sin 6 8^{\circ} = \frac{h}{5.2} \Rightarrow h = 5.2 sin 6 8^{\circ} \approx 4.82$ m.

b) $cos 6 8^{\circ} = \frac{d}{5.2} \Rightarrow d = 5.2 cos 6 8^{\circ} \approx 1.95$ m.

c) Now the opposite side (4.5) and the hypotenuse (5.2) are known: $sin θ = \frac{4.5}{5.2} \Rightarrow θ = sin^{- 1} (\frac{4.5}{5.2}) \approx 59. 9^{\circ}$

Check of a) and b): $4.8 2^{2} + 1.9 5^{2} \approx 27.0 = 5. 2^{2}$ .

Example 3 Hard

Height of a tower from two observations

From a point $P$ on horizontal ground, the angle of elevation to the top of a tower is $3 2^{\circ}$ . When you move 50 m in a straight line towards the base of the tower, to a point $Q$ , the angle of elevation becomes $5 1^{\circ}$ .

a) Calculate the height $h$ of the tower.

b) Calculate the horizontal distance from $P$ to the base of the tower.

c) From the top of the tower you can see a point $R$ on the ground on the other side of the tower, 80 m from its base. Calculate the angle of depression of $R$ from the top.

Solution

Let $x$ be the horizontal distance from $Q$ to the base of the tower. There are two right-angled triangles with the same vertical leg $h$ :

at $Q$ : $tan 5 1^{\circ} = \frac{h}{x}$ , so $x = \frac{h}{tan 5 1 ^{\circ}}$ ;
at $P$ : $tan 3 2^{\circ} = \frac{h}{x + 50}$ , so $x + 50 = \frac{h}{tan 3 2 ^{\circ}}$ .

a) Subtracting the two distances, $50 = \frac{h}{tan 3 2 ^{\circ}} - \frac{h}{tan 5 1 ^{\circ}}$ , from which $h = \frac{50}{\frac{1}{tan 3 2 ^{\circ}} - \frac{1}{tan 5 1 ^{\circ}}} \approx 63.2 m$

b) The distance from $P$ to the base is $\frac{h}{tan 3 2 ^{\circ}} \approx 101$ m. Check: that from $Q$ is $\frac{h}{tan 5 1 ^{\circ}} \approx 51.2$ m and $101.2 - 51.2 = 50$ m.

c) In the triangle formed by the tower and the ground up to $R$ , the angle of depression is equal to the angle at $R$ (alternate interior angles between parallel lines): $tan α = \frac{63.2}{80} \Rightarrow α = tan^{- 1} (\frac{63.2}{80}) \approx 38. 3^{\circ}$

In the exam: marks are given for drawing the two triangles with the common height, setting up the two equations with tangent, and justifying that the angle of depression is measured from the horizontal. You can solve the system directly on the GDC, but write down the equations you have set up.

Common mistakes

Summary

For an acute angle $θ$ : $sin θ = \frac{opp}{hyp}$ , $cos θ = \frac{adj}{hyp}$ , $tan θ = \frac{opp}{adj}$ (SOH-CAH-TOA rule).
To find a side, choose the ratio that contains the data and the unknown and rearrange; to find an angle, use $sin^{- 1}$ , $cos^{- 1}$ or $tan^{- 1}$ .
Solving a right-angled triangle means finding all its sides and angles: Pythagoras for the sides, and the two acute angles add up to $9 0^{\circ}$ .
Angles of elevation and depression are measured from the observer's horizontal; they are equal when seen from above and from below.
In problems with diagrams, look for the hidden right-angled triangle; in 3D, calculate step by step with flat triangles.
Check that the GDC is in degrees and round only at the end, to 3 significant figures.

Keep studying

Maths classes in Aldaia and online

Want to work on it together?

Free first consultation: we look at your level and your exam, and build a plan. Álvaro replies on WhatsApp.

Message us on WhatsApp