Angles and Triangles – Trigonometric Ratios in a Right Triangle

In a right-angled triangle, the lengths of the sides are linked to the acute angles by trigonometric ratios.

These ratios are used throughout geometry, physics and many real-world applications whenever we want to connect angles and lengths.

For an acute angle \(A\) in a right triangle, we label the sides relative to angle \(A\):

• Opposite: the side directly across from angle \(A\).
• Adjacent: the side next to angle \(A\), but not the hypotenuse.
• Hypotenuse: the longest side, opposite the right angle.

Only the position of the side matters: “opposite” and “adjacent” are always defined with respect to the angle we are working with.

If we change the angle, these labels may change, but the hypotenuse remains the same.

A common memory aid is the mnemonic SOH-CAH-TOA:

Worked Example – Solving Right Triangles

For each triangle, our goal is to find all unknown angles and sides. We combine trigonometric ratios with Pythagoras’ theorem, and we use inverse trig functions when solving for angles.

(a) Triangle \( \triangle ABC \)

Step 1 – Find the remaining acute angle

The acute angles in a right triangle are complementary:

$$ \theta = \angle A = 90^\circ – 41.53^\circ = 48.47^\circ $$

Step 2 – Find the hypotenuse \(AC\)

Relative to angle \(C\), side \(BC\) is adjacent and side \(AC\) is the hypotenuse. So we use cosine.

$$ \cos(41.53^\circ) = \frac{BC}{AC} $$

$$ AC = \frac{BC}{\cos(41.53^\circ)}
= \frac{4.66}{\cos(41.53^\circ)}
\approx 6.22\text{ cm (3 s.f.)} $$

Step 3 – Find the third side \(AB\)

$$ AB = \sqrt{AC^2 – BC^2}
= \sqrt{6.22^2 – 4.66^2}
\approx 4.13\text{ cm (3 s.f.)} $$

We now know both acute angles and all three sides. The triangle is completely solved.

(b) Triangle \( \triangle PQR \)

$$ RQ = \sqrt{PR^2 – PQ^2}
= \sqrt{6.97^2 – 4.87^2}
\approx 8.50\text{ cm (3 s.f.)} $$

Step 2 – Find angle at \(P\)

Using tangent, because we now know the opposite and adjacent sides relative to \(\angle P\):

$$ \tan(\angle PQR) = \frac{PR}{PQ} = \frac{6.97}{4.87} $$

$$ \angle PQR = \tan^{-1}\left(\frac{6.97}{4.87}\right)
\approx 55.1^\circ\text{ (3 s.f.)} $$

Step 3 – Find the remaining acute angle

$$ \angle PRQ = 90^\circ – 55.1^\circ \approx 34.9^\circ $$

In both parts, we begin with relationships that do not require inverse trig (Pythagoras or a direct trig equation), and then use inverse trig only when we need an angle value.

Angles of Elevation and Depression

Many applications of right-triangle trigonometry come from line-of-sight problems. We describe sight lines using angles of elevation and angles of depression.

• Angle of elevation – the angle formed when looking upwards from a horizontal line.
• Angle of depression – the angle formed when looking downwards from a horizontal line.

The horizontal line is usually taken at the observer’s eye level. The sight line and the horizontal line form a right triangle with the vertical height of the object being observed.

This allows us to translate angle-of-depression problems into standard right triangles on the ground.

Worked Example – Tree and Kite

(a) Finding the height of a tree

Emma stands 15 m from a tree. The angle of elevation to the top of the tree is \(40^\circ\). Her eye level is 1.42 m above the ground. Find the total height of the tree.

Solution

The vertical side of the right triangle from Emma’s eye to the top of the tree has length \(h\). The horizontal side is 15 m.

$$ \tan 40^\circ = \frac{h}{15} $$

$$ h = 15 \tan 40^\circ \approx 12.6\text{ m} $$

This is the height above Emma’s eye level. The total height of the tree is therefore

$$ \text{height of tree} = 12.6 + 1.42
\approx 14.0\text{ m (3 s.f.)}. $$

(b) Finding an angle of elevation to a kite

Frank stands on the opposite side of the tree. His eye level is 1.8 m, and the tree height is still approximately 14.0 m.

A kite is stuck at the top of the tree and the string from Frank’s hand to the kite is 16 m long.

Find the angle of elevation as Frank looks up at his kite.

Solution

The vertical distance from Frank’s eyes to the kite is the difference in heights:

$$ \text{opposite} = 14.0 – 1.8 = 12.2\text{ m (approximately)} $$

The hypotenuse is the length of the kite string, 16 m. Using the sine ratio:

$$ \sin \theta = \frac{12.206}{16} $$

$$ \theta = \sin^{-1}\left(\frac{12.206}{16}\right)
\approx 49.7^\circ\text{ (3 s.f.)}. $$

Non-Right Triangles and the Sine Rule

So far we have worked mainly with right triangles. However, many real-life triangles are not right-angled.

We can still connect angles and sides in these triangles using the Sine Rule.

Imagine drawing several different triangles:

• First, draw a scalene obtuse triangle \( \triangle ABC \) (no right angle).
• Measure all three side lengths \(a, b, c\) and all three angles \(\hat{A}, \hat{B}, \hat{C}\).
• Then calculate each of the following ratios, correct to 3 significant figures:

$$ \frac{\sin \hat{A}}{a},\quad
\frac{\sin \hat{B}}{b},\quad
\frac{\sin \hat{C}}{c}. $$

Next, repeat the same process for a scalene acute triangle \( \triangle DEF \) (again, no right angle).

Measure all the sides and angles, then calculate the ratios

$$ \frac{DE}{\sin \hat{F}},\quad
\frac{EF}{\sin \hat{D}},\quad
\frac{DF}{\sin \hat{E}}. $$

Finally, try this with a right-angled triangle as well. In each case, you should notice that all three ratios are (up to rounding) equal.

This experimental evidence leads to a general rule that works for any triangle, not just right-angled ones.

The Sine Rule – Statement and Use

When solving problems you will normally use only two of these three fractions to set up an equation.

• When solving for a side length, it is usually more convenient to use the version with the side in the numerator, for example \( \displaystyle \frac{a}{\sin \hat{A}} = \frac{b}{\sin \hat{B}} \).

• When solving for an angle, many students find it easier to rearrange so the sine of the unknown angle is on the left, for example \( \displaystyle \sin \hat{A} = \frac{a \sin \hat{B}}{b} \).

Worked Example – Finding an Angle with the Sine Rule

In triangle \( \triangle DEF \), we know:

• \(DE = 12\text{ cm}\)
• \(EF = 14\text{ cm}\)
• \(\widehat{DEF} = 45^\circ\)

Draw a labelled diagram and find the size of angle \(\widehat{EFD}\) (at vertex \(F\)) to the nearest degree.

Solution

Let \(\widehat{EFD} = \theta\). Then side 14 cm is opposite \(\theta\), and side 12 cm is opposite the given angle \(45^\circ\).

$$ \frac{\sin 45^\circ}{14}
= \frac{\sin \theta}{12}. $$

Rearranging to solve for \(\sin \theta\):

$$ \sin \theta = \frac{12 \sin 45^\circ}{14}. $$

Evaluate this using a calculator (ensuring degree mode):

$$ \sin \theta \approx \frac{12 \times \sin 45^\circ}{14}
\approx 0.6054\dots $$

Then use the inverse sine function:

$$ \theta = \sin^{-1}(0.6054\dots)
\approx 37.3^\circ. $$

To the nearest degree,

$$ \widehat{EFD} \approx 37^\circ. $$

Notice that we kept the exact calculator value for \(\sin \theta\) until the final step to reduce rounding errors.

Bearings and Navigation

A bearing is an angle measured clockwise from North. Bearings are usually written as a three-digit number, for example \(042^\circ\) or \(315^\circ\).

On a compass diagram:

• North corresponds to \(000^\circ\) or \(360^\circ\).
• East is \(090^\circ\), South is \(180^\circ\), West is \(270^\circ\).
• Intermediate directions (NE, SE, SW, NW, etc.) lie between these key bearings.

Worked Example – Bearings and the Sine Rule

A ship \(S\) is located off the coast so that:

• The bearing of \(S\) from port \(A\) is \(120^\circ\).
• The bearing of \(S\) from port \(B\) is \(042^\circ\).
• Port \(A\) is directly north of port \(B\).
• The distance between ports \(A\) and \(B\) is 15.2 miles.

Find the distance from the ship to each port.

Solution

Step 1: Identify the triangle

Draw a vertical segment \(AB\) with \(A\) above \(B\). From \(A\), the line of sight to \(S\) makes a \(120^\circ\) bearing, so the interior angle at \(A\) in triangle \(ABS\) is

$$ \widehat{A} = 180^\circ – 120^\circ = 60^\circ. $$

From \(B\), the bearing to \(S\) is \(042^\circ\). Measured clockwise from North, this forms an interior angle at \(B\) of

$$ \widehat{B} = 180^\circ – (42^\circ + 60^\circ) = 78^\circ. $$

Therefore the remaining angle at \(S\) is \(\widehat{S} = 180^\circ – 60^\circ – 78^\circ = 42^\circ\).

Step 2: Apply the Sine Rule

Let \(AS\) and \(BS\) be the distances from the ship to ports \(A\) and \(B\).
Side \(AB = 15.2\) miles is opposite angle \(\widehat{S} = 42^\circ\).

$$ \frac{AS}{\sin 42^\circ}
= \frac{15.2}{\sin 78^\circ}, \qquad
\frac{BS}{\sin 60^\circ}
= \frac{15.2}{\sin 78^\circ}. $$

$$ AS = \frac{\sin 42^\circ}{\sin 78^\circ} \times 15.2
\approx 10.4\text{ miles (3 s.f.)}, $$

$$ BS = \frac{\sin 60^\circ}{\sin 78^\circ} \times 15.2
\approx 13.5\text{ miles (3 s.f.)}. $$

Triangulation – Measuring Distance Indirectly

Triangulation is a technique used by surveyors to determine distances that are difficult to measure directly. By measuring:

• the distance between two landmarks, and
• the angles formed with a third point,

we create a triangle whose sides and angles we can calculate using trigonometry.

Repeating this process with several triangles allows us to build up a chain of distances across a landscape.

Worked Example – Triangulation Around a Lake

A lake has three docks at points \(B\), \(C\) and \(D\). The distance \(AB\) along a nearby highway is known to be 870 m.

Surveyors measure angles around the lake as shown in the diagram (for example \(\widehat{ABC} = 55^\circ\), \(\widehat{CBD} = 68^\circ\), \(\widehat{BDC} = 82^\circ\)).

(a) Use triangulation to find the distances \(BC\) and \(BD\).
(b) Nils rows at a speed of 1.5 m/s, starting from dock \(B\). How much longer will it take him to row to the further dock compared with the nearer one?

Solution

(a) First focus on triangle \( \triangle ABC \). We know side \(AB = 870\text{ m}\) and angles \(\widehat{A} = 55^\circ\), \(\widehat{B} = 57^\circ\) (so \(\widehat{C} = 180^\circ – 55^\circ – 57^\circ = 68^\circ\)).

Use the Sine Rule to find \(BC\).

$$ \frac{BC}{\sin 55^\circ}
= \frac{870}{\sin 57^\circ}, $$

$$ BC = \frac{870 \sin 55^\circ}{\sin 57^\circ}
\approx 950\text{ m (3 s.f.)}. $$

Next consider triangle \( \triangle BCD \) where we now know \(BC\), and the angles at \(C\) and \(D\) are \(61^\circ\) and \(82^\circ\) respectively.

Angle at \(B\) in this triangle is \(\widehat{B} = 180^\circ – 61^\circ – 82^\circ = 37^\circ\).

$$ \frac{BD}{\sin 61^\circ}
= \frac{BC}{\sin 82^\circ}, $$

$$ BD = \frac{BC \sin 61^\circ}{\sin 82^\circ}
\approx 962\text{ m (3 s.f.)}. $$

(b) The extra distance Nils must row to reach the further dock instead of the nearer one is \(BD – BC \approx 962 – 950 = 12\text{ m}\).

At 1.5 m/s, the extra time is

$$ t = \frac{\text{distance}}{\text{speed}}
= \frac{12}{1.5}
\approx 8.0\text{ s}. $$

The Ambiguous Case of the Sine Rule

In some situations we know two sides and a non-included angle (SSA). If the unknown angle lies opposite the longer of the two known sides, there may be two different triangles that fit the same data.

Consider a triangle where the known side opposite the known angle has length 3.7 cm, another side is 9.2 cm, and the given angle at one end is \(19^\circ\).

When we use the Sine Rule to find the unknown angle, the equation

$$ \sin \hat{A} = \frac{3.7 \sin 19^\circ}{9.2} $$

may give a sine value that corresponds to two possible angles:

\(\hat{A}\) and \(180^\circ – \hat{A}\).

Geometrically, we can draw two different triangles by “swinging” the unknown side to the other side of the line, both keeping the same side lengths and given angle.

A useful way to explore this is:

• Use the Sine Rule to find an angle in the first triangle configuration.
• Draw the alternate configuration where the longer side lies on the opposite side of the known base, and check that the new angle also satisfies the Sine Rule.
• Overlay the two diagrams to see how the possible triangles are related.

Additional information (such as whether the angle is acute or obtuse, or whether a point lies above/below a certain line) may allow us to decide which of the two triangles is physically meaningful.

Conceptual questions:

• Why does the Sine Rule sometimes produce two possible angle values?
• In which configurations (relative side lengths and known angle) do we get 0, 1, or 2 possible triangles?
• Why does the Sine Rule not always have just one solution in SSA cases, while the Cosine Rule does?

Area of a Triangle Using Trigonometry

Consider a triangle (ABC) with side lengths (a, b, c), where side (a) is opposite angle (A), side (b) is opposite angle (B), and side (c) is opposite angle (C). Draw the height (h) from vertex (C) perpendicular to side (AB) (which has length (c)).

Step 1 – Area in terms of base and height

The usual formula for the area of a triangle is

$$ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} c h. $$

Step 2 – Express the height using sine

Look at the right triangle formed by the height (h), side (a) and part of side (c).

The height (h) is opposite angle (B) and side (a) is the hypotenuse, so

$$ \sin B = \frac{h}{a} \quad \Rightarrow \quad h = a \sin B. $$

Step 3 – Substitute into the area formula

This is a powerful result: the area of any triangle can be found from two sides and the included angle.

Worked Example – Area from Two Sides and an Included Angle

A surveyor triangulating a region uses a tree on the opposite side of a river as a reference point. He measures:

• \(\angle ADC = 34^\circ\),
• \(DC = 47\text{ m}\),
• \(\angle C = 27^\circ\).

Calculate the area of triangle \(ACD\), which represents the region of interest.

Solution

Step 1 – Find the third angle

$$ A = 180^\circ – 34^\circ – 27^\circ = 119^\circ. $$

Step 2 – Use the Sine Rule to find the remaining side

We want side (AD), opposite angle (C = 27^\circ). Using the Sine Rule:

$$ \frac{AD}{\sin 27^\circ} = \frac{47}{\sin 119^\circ}. $$

$$ AD = \frac{47 \sin 27^\circ}{\sin 119^\circ}. $$

Step 3 – Find the area of triangle (ACD)

Use the area formula with two sides and the included angle (D = 34^\circ):

$$ \text{Area} = \frac{1}{2} \times AD \times DC \times \sin D. $$

Substitute \(AD\) from the previous step and \(DC = 47\text{ m}\):

$$ \text{Area} = \frac{1}{2} \times \frac{47 \sin 27^\circ}{\sin 119^\circ} \times 47 \times \sin 34^\circ. $$

Keeping full accuracy on the calculator until the final step,

$$ \text{Area} \approx 321\text{ m}^2 \quad (3\ \text{s.f.}). $$

Notice how we combined the Sine Rule (to find a missing side) with the area formula (to find the area).

The Cosine Rule

Sometimes we know all three side lengths of a triangle, or we know two sides and the included angle, but we do not know enough information to use the Sine Rule directly. In these cases we use the Cosine Rule.

The Cosine Rule is a generalisation of Pythagoras’ theorem:

• When \(A = 90^\circ\), \( \cos A = 0 \) and the formula reduces to \(a^2 = b^2 + c^2\).
• When \(A < 90^\circ\), \( \cos A > 0 \) and \(a^2 < b^2 + c^2\).
• When \(A > 90^\circ\), \( \cos A < 0 \) and \(a^2 > b^2 + c^2\).

This helps us interpret whether a triangle is acute or obtuse based on its side lengths alone.

Worked Example – Using the Cosine Rule (and Sine Rule)

A surveyor of a lake measures:

• \(AC = 225\text{ m}\),
• \(AB = 290\text{ m}\),
• \(\angle BAC = 72^\circ\).

Find:

1. the distance \(BC\);
2. the size of angle \(C\).

Solution

(a) – Find (BC)

We know two sides and the included angle at (A), so use the Cosine Rule with (BC) opposite angle (A):

$$ BC^2 = AB^2 + AC^2 – 2(AB)(AC)\cos \angle BAC. $$

Substitute the values:

$$ BC^2 = 290^2 + 225^2 – 2 \times 290 \times 225 \cos 72^\circ. $$

$$ BC \approx 307\text{ m} \quad (3\ \text{s.f.}). $$

(b) – Find angle (C)

Now that we know all three sides, we can use the Sine Rule to find angle (C):

$$ \frac{\sin C}{AB} = \frac{\sin A}{BC}
\quad \Rightarrow \quad
\frac{\sin C}{290} = \frac{\sin 72^\circ}{307}. $$

$$ \sin C = \frac{290 \sin 72^\circ}{307}. $$

$$ C = \sin^{-1}\!\left(\frac{290 \sin 72^\circ}{307}\right) \approx 63.9^\circ. $$

We have now determined both an unknown side and an unknown angle by combining the Cosine Rule and the Sine Rule.

Arc Length and Area of a Sector

Sometimes we are interested in only a portion of a circle:

• an arc, which is part of the circumference, or
• a sector, which is like a “slice” of the circle bounded by two radii and an arc.

If a circle has radius (r) and central angle (\theta) (measured in degrees), then the arc and sector are some fraction of the whole circle, proportional to (\theta / 360^\circ).

These formulas are used frequently in problems involving circular tracks, fields, wheels, or any situation where we only need part of a circle.

Worked Example – Arc Length, Chord Length, and Segment Area

A city park with a circular perimeter contains several sidewalks represented by the sides of triangle (ABC).

The sidewalk from the centre (A) to point (C) has length (AC = 85\text{ m}). The central angle (\angle BAC) is (75^\circ).

1. Find how much longer the arc \(BC\) is than the straight path (chord) \(BC\).
2. Find the area of the region between the chord \(BC\) and the arc \(BC\) (a circular segment).

Solution

(a) – Compare arc length and chord length

The radius of the circle is (r = 85\text{ m}), and the central angle is (\theta = 75^\circ).

Arc length:

$$ \widehat{BC} = \frac{75^\circ}{360^\circ} \times 2\pi (85). $$

Chord length (BC) can be found using the Sine Rule in triangle (ABC). Since (AB = AC = 85\text{ m}), triangle (ABC) is isosceles and the base angles at (B) and (C) are

$$ \angle B = \angle C = \frac{180^\circ – 75^\circ}{2} = 52.5^\circ. $$

Using the Sine Rule:

$$ \frac{BC}{\sin 75^\circ} = \frac{85}{\sin 52.5^\circ}
\quad \Rightarrow \quad
BC = \frac{85 \sin 75^\circ}{\sin 52.5^\circ}. $$

Evaluating both lengths:

$$ \widehat{BC} \approx 111.265\text{ m}, \qquad
BC \approx 103.489\text{ m}. $$

$$ 111.265 – 103.489 \approx 7.78\text{ m}. $$

So the arc is about 7.78 m longer than the chord.

(b) – Area of the circular segment

The segment area is the difference between the sector area and the area of triangle (ABC).

Sector area:

$$ A_{\text{sector}} = \frac{75^\circ}{360^\circ} \times \pi \times 85^2. $$

Triangle area (using the formula with two sides and included angle):

$$ A_{\text{triangle}} = \frac{1}{2} \times 85 \times 85 \times \sin 75^\circ. $$

Segment area:

$$ A_{\text{segment}} = A_{\text{sector}} – A_{\text{triangle}}. $$

Using a calculator and keeping full accuracy until the final step,

$$ A_{\text{segment}} \approx 1240\text{ m}^2. $$

This example shows how circular formulas, the Sine Rule, and the triangle area formula all work together in geometry problems involving arcs and chords.