## Sine, Cosine and Tangent of Acute Angles in Right-angled Triangles

 5.1 Sine, Cosine and Tangent of Acute Angles in Right-angled Triangles
 For a right-angled triangle, a) The hypotenuse is the longest side which is opposite the $$90{^\circ}$$ angle. b) The adjacent side and the opposite side change based on the position of the referred acute angle.
 Acute Angles in Right-Angled Triangles Given a fixed acute angle in right-angled triangles of different sizes: a) The ratio of the length of the opposite side to the hypotenuse is a constant. b) The ratio of the length of the adjacent side to the hypotenuse is a constant. c) The ratio of the length of the opposite side to the length of the adjacent side is a constant.
 Sine, Cosine and Tangent $$\text{Sine}=\dfrac{\text{Opposite side}}{\text{Hypotenuse}}$$ $$\text{Cosine}=\dfrac{\text{Adjacent side}}{\text{Hypotenuse}}$$ $$\text{Tangent}=\dfrac{\text{Opposite side}}{\text{Adjacent side}}$$
 Changing the Size of the Angles The larger the size of the acute angle: a) the larger the value of sine and its value approaches $$1$$. ​b) the smaller the value of cosine and its value approaches $$0$$. c) the larger the value of tangent.
 Example The following diagram shows a right-angled triangle. Calculate the value of: a) the length $$PR$$ \begin{aligned} PR&= \sqrt{15^2 + 8^2} \\\\&=\sqrt{289} \\\\&= 17 \text{ cm}. \end{aligned} b) $$\text{sin }\angle PRQ$$ \begin{aligned} \text{sin}&=\dfrac{\text{opposite}}{\text{hypotenuse}} \\\\&=\dfrac{PQ}{PR} \\\\&=\dfrac{15}{17}. \end{aligned} c) $$\text{cos }\angle PRQ$$ \begin{aligned} \text{cos}&=\dfrac{\text{adjacent}}{\text{hypotenuse}} \\\\&=\dfrac{QR}{PR} \\\\&=\dfrac{8}{17}. \end{aligned} d) $$\text{tan }\angle QPR$$ \begin{aligned} \text{tan}&=\dfrac{\text{opposite}}{\text{adjacent}} \\\\&=\dfrac{QR}{PQ} \\\\&=\dfrac{8}{15}. \end{aligned}
 Relationship between sine, cosine and tangent: Example Given that $$\text{sin }\theta=0.6$$ and $$\text{cos }\theta=0.8$$. What is the value of $$\text{tan }\theta$$? \begin{aligned} \text{tan } \theta&=\dfrac{\text{sin }\theta}{\text{cos }\theta} \\\\&=\dfrac{0.6}{0.8} \\\\&=\dfrac{3}{4} \\\\&=0.75. \end{aligned}
 Example The diagram below shows a right-angled triangle $$PQR$$. Given that $$PR= 20 \text{ cm}$$ and $$\text{sin } \angle QPR= \dfrac{3}{5}$$. a) Determine the length of $$QR$$. Noted that $$\text{sin } \angle QPR= \dfrac{3}{5}$$. So, \begin{aligned} \text{sin } \angle QPR &=\dfrac{3}{5} \\\\ \dfrac{QR}{PR}&=\dfrac{3}{5} \\\\ \dfrac{QR}{20}&=\dfrac{3}{5} \\\\ QR&=\dfrac{3}{5}\times20 \\\\ QR&= 12 \text{ cm}. \end{aligned} b) Calculate the value of $$\text{cos }\angle QPR$$. First, we need to calculate the length of $$PQ$$. \begin{aligned} PQ&=\sqrt{PR^2-QR^2} \\\\&= \sqrt{20^2-12^2} \\\\&= \sqrt{256} \\\\&= 16 \text{ cm}. \\\\\end{aligned} Hence, \begin{aligned} \text{cos } \angle QPR&=\dfrac{PQ}{PR} \\\\&=\dfrac{16}{20} \\\\&=\dfrac{4}{5}. \end{aligned}

The values of sine, cosine and tangent of $$30^\circ$$, $$45^\circ$$ and $$60^\circ$$ angles without using a calculator:

 Ratio/ Angle $$30^\circ$$ $$45^\circ$$ $$60^\circ$$ $$\text{sin }\theta$$ $$\dfrac{1}{2}$$ $$\dfrac{\sqrt{3}}{2}$$ $$\dfrac{1}{\sqrt{2}}$$ $$\text{cos }\theta$$ $$\dfrac{\sqrt{3}}{2}$$ $$\dfrac{1}{2}$$ $$\dfrac{1}{\sqrt{2}}$$ $$\text{tan }\theta$$ $$\dfrac{1}{\sqrt{3}}$$ $$\sqrt{3}$$ $$1$$

 Example Calculate $$2\text{ cos }60^\circ+\text{tan }45^\circ$$. \begin{aligned}&\space2\text{ cos }60^\circ+\text{tan }45^\circ \\\\&=2(\dfrac{1}{2})+1 \\\\&=2. \end{aligned}

Unit of measure for angles:

• Angles are measured in the unit of degrees ( $$^\circ$$ ), minutes ( $$'$$ ) and seconds ( $$''$$ ).

Noted that,

\begin{aligned}1^\circ&=60' \\\\1'&=60'' \end{aligned}

 Example Convert $$58.1^\circ$$ to degrees and minutes. \begin{aligned} 58.1^\circ&=58^\circ+0.1^\circ \\\\&=58^\circ+(0.1\times60)' \\\\&=57^\circ+6' \\\\&=58^\circ\,6'. \end{aligned}

## Sine, Cosine and Tangent of Acute Angles in Right-angled Triangles

 5.1 Sine, Cosine and Tangent of Acute Angles in Right-angled Triangles
 For a right-angled triangle, a) The hypotenuse is the longest side which is opposite the $$90{^\circ}$$ angle. b) The adjacent side and the opposite side change based on the position of the referred acute angle.
 Acute Angles in Right-Angled Triangles Given a fixed acute angle in right-angled triangles of different sizes: a) The ratio of the length of the opposite side to the hypotenuse is a constant. b) The ratio of the length of the adjacent side to the hypotenuse is a constant. c) The ratio of the length of the opposite side to the length of the adjacent side is a constant.
 Sine, Cosine and Tangent $$\text{Sine}=\dfrac{\text{Opposite side}}{\text{Hypotenuse}}$$ $$\text{Cosine}=\dfrac{\text{Adjacent side}}{\text{Hypotenuse}}$$ $$\text{Tangent}=\dfrac{\text{Opposite side}}{\text{Adjacent side}}$$
 Changing the Size of the Angles The larger the size of the acute angle: a) the larger the value of sine and its value approaches $$1$$. ​b) the smaller the value of cosine and its value approaches $$0$$. c) the larger the value of tangent.
 Example The following diagram shows a right-angled triangle. Calculate the value of: a) the length $$PR$$ \begin{aligned} PR&= \sqrt{15^2 + 8^2} \\\\&=\sqrt{289} \\\\&= 17 \text{ cm}. \end{aligned} b) $$\text{sin }\angle PRQ$$ \begin{aligned} \text{sin}&=\dfrac{\text{opposite}}{\text{hypotenuse}} \\\\&=\dfrac{PQ}{PR} \\\\&=\dfrac{15}{17}. \end{aligned} c) $$\text{cos }\angle PRQ$$ \begin{aligned} \text{cos}&=\dfrac{\text{adjacent}}{\text{hypotenuse}} \\\\&=\dfrac{QR}{PR} \\\\&=\dfrac{8}{17}. \end{aligned} d) $$\text{tan }\angle QPR$$ \begin{aligned} \text{tan}&=\dfrac{\text{opposite}}{\text{adjacent}} \\\\&=\dfrac{QR}{PQ} \\\\&=\dfrac{8}{15}. \end{aligned}
 Relationship between sine, cosine and tangent: Example Given that $$\text{sin }\theta=0.6$$ and $$\text{cos }\theta=0.8$$. What is the value of $$\text{tan }\theta$$? \begin{aligned} \text{tan } \theta&=\dfrac{\text{sin }\theta}{\text{cos }\theta} \\\\&=\dfrac{0.6}{0.8} \\\\&=\dfrac{3}{4} \\\\&=0.75. \end{aligned}
 Example The diagram below shows a right-angled triangle $$PQR$$. Given that $$PR= 20 \text{ cm}$$ and $$\text{sin } \angle QPR= \dfrac{3}{5}$$. a) Determine the length of $$QR$$. Noted that $$\text{sin } \angle QPR= \dfrac{3}{5}$$. So, \begin{aligned} \text{sin } \angle QPR &=\dfrac{3}{5} \\\\ \dfrac{QR}{PR}&=\dfrac{3}{5} \\\\ \dfrac{QR}{20}&=\dfrac{3}{5} \\\\ QR&=\dfrac{3}{5}\times20 \\\\ QR&= 12 \text{ cm}. \end{aligned} b) Calculate the value of $$\text{cos }\angle QPR$$. First, we need to calculate the length of $$PQ$$. \begin{aligned} PQ&=\sqrt{PR^2-QR^2} \\\\&= \sqrt{20^2-12^2} \\\\&= \sqrt{256} \\\\&= 16 \text{ cm}. \\\\\end{aligned} Hence, \begin{aligned} \text{cos } \angle QPR&=\dfrac{PQ}{PR} \\\\&=\dfrac{16}{20} \\\\&=\dfrac{4}{5}. \end{aligned}

The values of sine, cosine and tangent of $$30^\circ$$, $$45^\circ$$ and $$60^\circ$$ angles without using a calculator:

 Ratio/ Angle $$30^\circ$$ $$45^\circ$$ $$60^\circ$$ $$\text{sin }\theta$$ $$\dfrac{1}{2}$$ $$\dfrac{\sqrt{3}}{2}$$ $$\dfrac{1}{\sqrt{2}}$$ $$\text{cos }\theta$$ $$\dfrac{\sqrt{3}}{2}$$ $$\dfrac{1}{2}$$ $$\dfrac{1}{\sqrt{2}}$$ $$\text{tan }\theta$$ $$\dfrac{1}{\sqrt{3}}$$ $$\sqrt{3}$$ $$1$$

 Example Calculate $$2\text{ cos }60^\circ+\text{tan }45^\circ$$. \begin{aligned}&\space2\text{ cos }60^\circ+\text{tan }45^\circ \\\\&=2(\dfrac{1}{2})+1 \\\\&=2. \end{aligned}

Unit of measure for angles:

• Angles are measured in the unit of degrees ( $$^\circ$$ ), minutes ( $$'$$ ) and seconds ( $$''$$ ).

Noted that,

\begin{aligned}1^\circ&=60' \\\\1'&=60'' \end{aligned}

 Example Convert $$58.1^\circ$$ to degrees and minutes. \begin{aligned} 58.1^\circ&=58^\circ+0.1^\circ \\\\&=58^\circ+(0.1\times60)' \\\\&=57^\circ+6' \\\\&=58^\circ\,6'. \end{aligned}