Given , calculate all other trigonometric ratios.

Understand the Question Given , let and , where is a positive constant. Use the Pythagoras theorem to calculate the length of the opposite side. Use the side lengths to determine the remaining five trigonometric ratios: , , , , and . Step 1 · Find the Opposite Side Consider right-angled triangle with and . Given Let and for some positive number . By Pythagoras theorem in Step 2 · Calculate Other Trigonometric Ratios Using the side lengths: , , and Answer > Common Mistakes Swapping Opposite and Adjacent Sides: Ensure side is adjacent to angle and is opposite to angle . Reciprocal Shortcut: Notice that can be written directly without finding the third side first.

Introduction to Trigonometry | Exercise 8.1

Question 5

Given $\sec \theta = \dfrac{13}{12}$ , calculate all other trigonometric ratios.

Solution

Understand the Question

Given $\sec \theta = \dfrac{13}{12} = \dfrac{\text{Hypotenuse}}{\text{Adjacent}}$ , let $\text{Hypotenuse} = 13k$ and $\text{Adjacent} = 12k$ , where $k$ is a positive constant.
Use the Pythagoras theorem to calculate the length of the opposite side.
Use the side lengths to determine the remaining five trigonometric ratios: $\sin \theta$ , $\cos \theta$ , $\tan \theta$ , $\csc \theta$ , and $\cot \theta$ .

Step 1 · Find the Opposite Side

Consider right-angled triangle $\text{ABC}$ with $\angle \text{B} = 90^\circ$ and $\angle \text{A} = \theta$ . Diagram 1

Given $\sec \theta = \dfrac{\text{Hypotenuse}}{\text{Adjacent side}} = \dfrac{\text{AC}}{\text{AB}} = \dfrac{13}{12}$

Let $\text{AC} = 13k$ and $\text{AB} = 12k$ for some positive number $k$ .

By Pythagoras theorem in $\Delta \text{ABC}$

\begin{aligned} \text{AC}^2 &= \text{AB}^2 + \text{BC}^2 \\ (13k)^2 &= (12k)^2 + \text{BC}^2 \\ 169k^2 &= 144k^2 + \text{BC}^2 \\ \text{BC}^2 &= 169k^2 - 144k^2 \\ \text{BC}^2 &= 25k^2 \\ \text{BC} &= \sqrt{25k^2} = 5k \end{aligned}

Step 2 · Calculate Other Trigonometric Ratios

Using the side lengths: $\text{Opposite} = \text{BC} = 5k$ , $\text{Adjacent} = \text{AB} = 12k$ , and $\text{Hypotenuse} = \text{AC} = 13k$

$\sin \theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}} = \dfrac{\text{BC}}{\text{AC}} = \dfrac{5k}{13k} = \dfrac{5}{13}$

$\cos \theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}} = \dfrac{\text{AB}}{\text{AC}} = \dfrac{12k}{13k} = \dfrac{12}{13}$

$\tan \theta = \dfrac{\text{Opposite}}{\text{Adjacent}} = \dfrac{\text{BC}}{\text{AB}} = \dfrac{5k}{12k} = \dfrac{5}{12}$

$\csc \theta = \dfrac{\text{Hypotenuse}}{\text{Opposite}} = \dfrac{\text{AC}}{\text{BC}} = \dfrac{13k}{5k} = \dfrac{13}{5}$

$\cot \theta = \dfrac{\text{Adjacent}}{\text{Opposite}} = \dfrac{\text{AB}}{\text{BC}} = \dfrac{12k}{5k} = \dfrac{12}{5}$

Answer

$\sin \theta = \dfrac{5}{13}, \quad \cos \theta = \dfrac{12}{13}, \quad \tan \theta = \dfrac{5}{12}, \quad \csc \theta = \dfrac{13}{5}, \quad \cot \theta = \dfrac{12}{5}$

Common Mistakes

Swapping Opposite and Adjacent Sides: Ensure side $\text{AB}$ is adjacent to angle $\theta$ and $\text{BC}$ is opposite to angle $\theta$ .
Reciprocal Shortcut: Notice that $\cos \theta = \dfrac{1}{\sec \theta} = \dfrac{12}{13}$ can be written directly without finding the third side first.

Question 5

More questions in Exercise 8.1