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.

	\(\begin{aligned} 58.1^\circ&=58^\circ+0.1^\circ \\\\&=58^\circ+(0.1\times60)' \\\\&=57^\circ+6' \\\\&=58^\circ\,6'. \end{aligned}\)