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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.  
     
  \(\begin{aligned} 58.1^\circ&=58^\circ+0.1^\circ \\\\&=58^\circ+(0.1\times60)' \\\\&=57^\circ+6' \\\\&=58^\circ\,6'. \end{aligned}\)  
 
The views expressed in the notes here are those of the contributor and don’t necessarily represent the views of pandai.

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.  
     
  \(\begin{aligned} 58.1^\circ&=58^\circ+0.1^\circ \\\\&=58^\circ+(0.1\times60)' \\\\&=57^\circ+6' \\\\&=58^\circ\,6'. \end{aligned}\)  
 
The views expressed in the notes here are those of the contributor and don’t necessarily represent the views of pandai.
Chapter : Trigonometric Ratios
Topic : Solve problems involving sine, cosine and tangent
Form 3 Mathematics
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