Trigonometry: Finding Sides and Angles in Right-Angled Triangles

What the examiner is testing

The examiner is testing your ability to identify the correct trigonometric ratio (sine, cosine, or tangent) and apply it to calculate unknown side lengths or angles within a right-angled triangle. Marks are awarded for correctly setting up the equation, accurate calculation, and providing answers to the specified degree of accuracy.

The method

1. Identify the right angle: Locate the \(90^\circ\) angle in the triangle.
2. Identify the known angle: Find the angle (other than the right angle) that is either given or needs to be found.
3. Label the sides relative to the known angle:
   • Opposite (O): The side directly across from the known angle.
   • Adjacent (A): The side next to the known angle that is not the hypotenuse.
   • Hypotenuse (H): The longest side, always opposite the right angle.
4. Choose the correct trigonometric ratio: Use "SOH CAH TOA" to decide which ratio involves the two sides you are working with (one known, one unknown):
   • SOH: \(\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}\)
   • CAH: \(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)
   • TOA: \(\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}\)
5. Set up the equation: Substitute the known values into the chosen trigonometric ratio.
6. Solve the equation:
   • If finding a side: Rearrange the equation to isolate the unknown side.
   • If finding an angle: Use the inverse trigonometric function (\(\sin^{-1}\), \(\cos^{-1}\), or \(\tan^{-1}\)).
7. Calculate and state the answer: Use a calculator and round your answer to the required degree of accuracy (e.g., one decimal place, three significant figures). Include units for side lengths.

Worked example

Find the length of side \(x\) in the triangle below. Give your answer to 1 decimal place.

$$ \begin{array}{c} \text{Diagram: A right-angled triangle with angles } 90^\circ \text{ and } 35^\circ. \\ \text{The side opposite the } 35^\circ \text{ angle is } x. \\ \text{The hypotenuse is } 12 \text{ cm.} \end{array} $$

1. The right angle is marked.
2. The known angle is \(35^\circ\).
3. Relative to \(35^\circ\):
   • Opposite (O) = \(x\)
   • Hypotenuse (H) = \(12 \text{ cm}\)
   • Adjacent (A) is not involved.
4. We have O and H, so we use SOH: \(\sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}\).
5. Set up the equation:
   \(\sin(35^\circ) = \frac{x}{12}\)
6. Solve for \(x\):
   \(x = 12 \times \sin(35^\circ)\)
7. Calculate:
   \(x = 12 \times 0.573576...\)
   \(x = 6.8829...\)
   \(x = 6.9 \text{ cm}\) (to 1 decimal place)

Sanity check: The hypotenuse is \(12 \text{ cm}\). Side \(x\) is opposite an angle of \(35^\circ\), which is less than \(45^\circ\). This means \(x\) should be shorter than the adjacent side and certainly shorter than the hypotenuse. \(6.9 \text{ cm}\) is shorter than \(12 \text{ cm}\), so the answer is reasonable.

Worked example: a harder one

A ladder of length 5.5 m leans against a vertical wall. The base of the ladder is 1.8 m away from the wall on horizontal ground. Calculate the angle the ladder makes with the ground, to one decimal place.

$$ \begin{array}{c} \text{Diagram: A right-angled triangle formed by the wall, ground, and ladder.} \\ \text{The hypotenuse (ladder) is } 5.5 \text{ m.} \\ \text{The adjacent side (ground) is } 1.8 \text{ m.} \\ \text{The angle between the ladder and the ground is } \theta. \end{array} $$

Why the obvious first move fails (or isn't the most efficient): You might think about finding the height the ladder reaches up the wall first using Pythagoras. While possible, this adds an unnecessary step. The question asks for an angle, and we already have two sides that relate directly to that angle.

1. The right angle is formed between the wall and the ground.
2. The angle to be found is \(\theta\), the angle the ladder makes with the ground.
3. Relative to \(\theta\):
   • Adjacent (A) = \(1.8 \text{ m}\) (distance from wall)
   • Hypotenuse (H) = \(5.5 \text{ m}\) (length of ladder)
   • Opposite (O) is not involved.
4. We have A and H, so we use CAH: \(\cos(\theta) = \frac{\text{Adjacent}}{\text{Hypotenuse}}\).
5. Set up the equation:
   \(\cos(\theta) = \frac{1.8}{5.5}\)
6. Solve for \(\theta\):
   \(\theta = \cos^{-1}\left(\frac{1.8}{5.5}\right)\)
7. Calculate:
   \(\theta = \cos^{-1}(0.327272...)\)
   \(\theta = 71.014...\)
   \(\theta = 71.0^\circ\) (to 1 decimal place)

Practice

1. Find the length of side \(y\). Give your answer to 3 significant figures.
   $$ \begin{array}{c} \text{Diagram: A right-angled triangle with angles } 90^\circ \text{ and } 50^\circ. \\ \text{The side adjacent to the } 50^\circ \text{ angle is } y. \\ \text{The hypotenuse is } 8 \text{ cm.} \end{array} $$
2. Calculate the angle \(\alpha\). Give your answer to 1 decimal place.
   $$ \begin{array}{c} \text{Diagram: A right-angled triangle with angles } 90^\circ \text{ and } \alpha. \\ \text{The side opposite } \alpha \text{ is } 7 \text{ m.} \\ \text{The adjacent side to } \alpha \text{ is } 4 \text{ m.} \end{array} $$
3. A ramp is 4.5 m long and rises at an angle of \(15^\circ\) to the horizontal. How high does the ramp rise vertically? Give your answer to 2 decimal places.
4. In the diagram, calculate the perimeter of the triangle ABC. Give your answer to 3 significant figures.
   $$ \begin{array}{c} \text{Diagram: A right-angled triangle ABC, with the right angle at B.} \\ \text{Angle BAC is } 62^\circ. \\ \text{Side BC is } 10 \text{ cm.} \end{array} $$

Answers:
1. \(y = 8 \times \cos(50^\circ) = 5.14 \text{ cm}\) (3 s.f.)
2. \(\tan(\alpha) = \frac{7}{4} \implies \alpha = \tan^{-1}\left(\frac{7}{4}\right) = 60.3^\circ\) (1 d.p.)
3. \(\sin(15^\circ) = \frac{\text{height}}{4.5} \implies \text{height} = 4.5 \times \sin(15^\circ) = 1.16 \text{ m}\) (2 d.p.)
4. Working for Q4:
* We need sides AB and AC.
* To find AC (hypotenuse):
\(\sin(62^\circ) = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{10}{\text{AC}}\)
\(\text{AC} = \frac{10}{\sin(62^\circ)}\)
\(\text{AC} = \frac{10}{0.8829...}\)
\(\text{AC} = 11.325...\text{ cm}\)
* To find AB (adjacent):
\(\tan(62^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{10}{\text{AB}}\)
\(\text{AB} = \frac{10}{\tan(62^\circ)}\)
\(\text{AB} = \frac{10}{1.8807...}\)
\(\text{AB} = 5.317...\text{ cm}\)
* Perimeter = AB + BC + AC
Perimeter = \(5.317... + 10 + 11.325...\)
Perimeter = \(26.642...\)
Perimeter = \(26.6 \text{ cm}\) (3 s.f.)

The three mistakes that lose marks

1. Using the wrong ratio: Mixing up SOH, CAH, TOA. For example, using \(\cos\) when you should use \(\sin\).
   • Wrong Answer Example: If given the opposite side and the hypotenuse, but you incorrectly use \(\cos(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}}\), your calculation will be wrong from the start.
2. Calculator in the wrong mode: Your calculator must be in "DEG" (degrees) mode, not "RAD" (radians) or "GRAD" (gradians).
   • Wrong Answer Example: If finding \(\sin(30^\circ)\) in radian mode, you'd get approximately \(-0.988\), instead of \(0.5\).
3. Incorrectly rearranging the equation: Especially when the unknown is in the denominator.
   • Wrong Answer Example: If \(\tan(40^\circ) = \frac{15}{x}\), some students might incorrectly write \(x = 15 \times \tan(40^\circ)\) instead of \(x = \frac{15}{\tan(40^\circ)}\). This would give \(12.58...\) instead of \(17.87...\).

30-second recap

Trigonometry in right-angled triangles uses SOH CAH TOA to relate angles and side lengths. Label the sides (Opposite, Adjacent, Hypotenuse) relative to the known or unknown angle. Choose the correct ratio, set up the equation, and solve using inverse functions for angles.