In , right-angled at B, , . Determine : (i) , (ii) ,

Understand the Question In right triangle with right angle at , we are given and . First, find the hypotenuse using the Pythagoras theorem: Trigonometric ratios depend on which acute angle is being referenced: For , and . For , and . Part (i) Determine , Step 1 · Find Hypotenuse AC In right , right-angled at : By Pythagoras theorem Step 2 · Calculate and For angle : Answer > (i) , Part (ii) Determine , Step 1 · Calculate and For angle : Answer > (ii) , Common Mistakes Switching Opposite and Adjacent Sides: The opposite and adjacent sides depend on the angle considered. For , the opposite side is , whereas for , the opposite side is . Hypotenuse Identification: The hypotenuse is always the side opposite the angle (here, ), and it remains fixed regardless of whether finding ratios for or .

Introduction to Trigonometry | Exercise 8.1

Question 1

In $\Delta \text{ABC}$ , right-angled at B, $\text{AB} = 24\text{ cm}$ , $\text{BC} = 7\text{ cm}$ . Determine :

(i) $\sin A$ , $\cos A$

(ii) $\sin C$ , $\cos C$

Solution

Understand the Question

In right triangle $\text{ABC}$ with right angle at $\text{B}$ , we are given $\text{AB} = 24\text{ cm}$ and $\text{BC} = 7\text{ cm}$ .
First, find the hypotenuse $\text{AC}$ using the Pythagoras theorem: $\text{AC}^2 = \text{AB}^2 + \text{BC}^2$
Trigonometric ratios depend on which acute angle is being referenced:
- $\sin \theta = \dfrac{\text{Opposite}}{\text{Hypotenuse}}$
- $\cos \theta = \dfrac{\text{Adjacent}}{\text{Hypotenuse}}$
For $\angle \text{A}$ , $\text{Opposite} = \text{BC}$ and $\text{Adjacent} = \text{AB}$ .
For $\angle \text{C}$ , $\text{Opposite} = \text{AB}$ and $\text{Adjacent} = \text{BC}$ .

(i) Determine $\sin A$ , $\cos A$

Step 1 · Find Hypotenuse AC

In right $\Delta \text{ABC}$ , right-angled at $\text{B}$ : Diagram 1

By Pythagoras theorem

\begin{aligned} \text{AC}^2 &= \text{AB}^2 + \text{BC}^2 \\ &= (24\text{ cm})^2 + (7\text{ cm})^2 \\ &= 576\text{ cm}^2 + 49\text{ cm}^2 \\ &= 625\text{ cm}^2 \\ \text{AC} &= \sqrt{625\text{ cm}^2} = 25\text{ cm} \end{aligned}

Step 2 · Calculate $\sin A$ and $\cos A$

For angle $\text{A}$ :

$\text{Opposite side} = \text{BC} = 7\text{ cm}$
$\text{Adjacent side} = \text{AB} = 24\text{ cm}$
$\text{Hypotenuse} = \text{AC} = 25\text{ cm}$

\begin{aligned} \sin A &= \dfrac{\text{Opposite side}}{\text{Hypotenuse}} = \dfrac{\text{BC}}{\text{AC}} = \dfrac{7}{25} \\[0.6em] \cos A &= \dfrac{\text{Adjacent side}}{\text{Hypotenuse}} = \dfrac{\text{AB}}{\text{AC}} = \dfrac{24}{25} \end{aligned}

Answer

(i) $\sin A = \dfrac{7}{25}$ , $\cos A = \dfrac{24}{25}$

(ii) Determine $\sin C$ , $\cos C$

Step 1 · Calculate $\sin C$ and $\cos C$

For angle $\text{C}$ :

$\text{Opposite side} = \text{AB} = 24\text{ cm}$
$\text{Adjacent side} = \text{BC} = 7\text{ cm}$
$\text{Hypotenuse} = \text{AC} = 25\text{ cm}$

\begin{aligned} \sin C &= \dfrac{\text{Opposite side}}{\text{Hypotenuse}} = \dfrac{\text{AB}}{\text{AC}} = \dfrac{24}{25} \\[0.6em] \cos C &= \dfrac{\text{Adjacent side}}{\text{Hypotenuse}} = \dfrac{\text{BC}}{\text{AC}} = \dfrac{7}{25} \end{aligned}

Answer

(ii) $\sin C = \dfrac{24}{25}$ , $\cos C = \dfrac{7}{25}$

Common Mistakes

Switching Opposite and Adjacent Sides: The opposite and adjacent sides depend on the angle considered. For $\angle \text{A}$ , the opposite side is $\text{BC}$ , whereas for $\angle \text{C}$ , the opposite side is $\text{AB}$ .
Hypotenuse Identification: The hypotenuse is always the side opposite the $90^\circ$ angle (here, $\text{AC} = 25\text{ cm}$ ), and it remains fixed regardless of whether finding ratios for $\angle \text{A}$ or $\angle \text{C}$ .

Question 1

More questions in Exercise 8.1