Understand the Question Given , let and , where is a positive number. To evaluate and , we first find the adjacent side using the Pythagoras theorem. Step 1 · Find the Missing Side Consider right triangle right-angled at . Given Let and for some positive number . By Pythagoras theorem in Step 2 · Calculate Step 3 · Calculate Answer > Common Mistakes Pythagoras Error: Adding instead of subtracting . Remember that the hypotenuse is the longest side (). Ratio Swapping: Confusing with .

Introduction to Trigonometry | Exercise 8.1

Question 3

If $\sin A = \dfrac{3}{4}$ , calculate $\cos A$ and $\tan A$ .

Solution

Understand the Question

Given $\sin A = \dfrac{3}{4} = \dfrac{\text{Opposite}}{\text{Hypotenuse}}$ , let $\text{Opposite} = 3k$ and $\text{Hypotenuse} = 4k$ , where $k$ is a positive number.
To evaluate $\cos A = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$ and $\tan A = \dfrac{\text{Opposite}}{\text{Adjacent}}$ , we first find the adjacent side using the Pythagoras theorem.

Step 1 · Find the Missing Side

Consider right triangle $\text{ABC}$ right-angled at $\text{B}$ . Diagram 1

Given $\sin A = \dfrac{\text{Opposite side}}{\text{Hypotenuse}} = \dfrac{\text{BC}}{\text{AC}} = \dfrac{3}{4}$

Let $\text{BC} = 3k$ and $\text{AC} = 4k$ for some positive number $k$ .

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

\begin{aligned} \text{AC}^2 &= \text{AB}^2 + \text{BC}^2 \\ (4k)^2 &= \text{AB}^2 + (3k)^2 \\ 16k^2 &= \text{AB}^2 + 9k^2 \\ \text{AB}^2 &= 16k^2 - 9k^2 \\ \text{AB}^2 &= 7k^2 \\ \text{AB} &= \sqrt{7k^2} = k\sqrt{7} \end{aligned}

Step 2 · Calculate $\cos A$

\begin{aligned} \cos A &= \dfrac{\text{Adjacent side}}{\text{Hypotenuse}} \\[0.6em] &= \dfrac{\text{AB}}{\text{AC}} \\[0.6em] &= \dfrac{k\sqrt{7}}{4k} \\[0.6em] &= \dfrac{\sqrt{7}}{4} \end{aligned}

Step 3 · Calculate $\tan A$

\begin{aligned} \tan A &= \dfrac{\text{Opposite side}}{\text{Adjacent side}} \\[0.6em] &= \dfrac{\text{BC}}{\text{AB}} \\[0.6em] &= \dfrac{3k}{k\sqrt{7}} \\[0.6em] &= \dfrac{3}{\sqrt{7}} \end{aligned}

Answer

$\cos A = \dfrac{\sqrt{7}}{4}, \quad \tan A = \dfrac{3}{\sqrt{7}}$

Common Mistakes

Pythagoras Error: Adding $4^2 + 3^2 = 25$ instead of subtracting $4^2 - 3^2 = 7$ . Remember that the hypotenuse is the longest side ( $ext{Hypotenuse}^2 = ext{Opposite}^2 + ext{Adjacent}^2$ ).
Ratio Swapping: Confusing $\cos A = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$ with $\tan A = \dfrac{\text{Opposite}}{\text{Adjacent}}$ .

Question 3

More questions in Exercise 8.1