\(\textbf{Some History}\)

In the \(1800\)s, when India was still a British colony, a British surveyor stationed not far away from the foothills of the Himalayas sighted an extremely high peak in the distance. When a mathematician calculated its height based on measurements taken from a survey, he realised that it was the world's highest mountain. It was then named Mount Everest, in honour of Sir George Everest, the Surveyor-General of India.

How was the mountain's height measured?

\(\bullet\) Firstly, a point \(P\) on the ground away from the mountain was chosen.

\(\bullet\) Next, the distance \(d\) from \(P\) to the base of the mountain was measured as precisely as possible (this was extremely difficult to do due to hilly terrain).

\(\bullet\) The angle of elevation \(\theta\) of the summit from the point \(P\) was then measured.

\(\bullet\) Using trigonometry, the mountain's height was finally determined.

What, then, is trigonometry? Basically, it is the study of the relationship between angles and lengths.

\(\textbf{1. The Trigonometric Ratios}\)

**a. Defining the trigonometric ratios at acute angles**

Let \(\theta\) be an acute angle, i.e. \(0^{\circ} < \theta < 90^{\circ}\). Draw a right-angled triangle \(\triangle ABC\) with \(\angle A=\theta\) and \(\angle B=90^{\circ}\).

The **sine**, **cosine** and **tangent** ratios at the angle \(\theta\) are then defined as follows:

\[\sin \theta = \frac{\text{Opposite side}}{\text{Hypothenuse}}=\frac{BC}{AC};\]

\[\cos \theta = \frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{AB}{AC};\]

\[\tan \theta = \frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{BC}{AB}.\]

To remember these definitions more easily, memorise the mnemonic 'toa-cah-soh' or 'soh-cah-toa'.

**b. Exercises**

\(1\). Use the concept of similar triangles to show that the definitions of the three basic trigonometric ratios at an acute angle do not depend on the size of the right-angled triangle drawn.

\(2\). Show that, for acute \(\theta\), \(\cos(90^{\circ}-\theta)=\sin \theta\).

\(3\). Construct \(\triangle ABC\) such that \(\angle B=90^{\circ}, AB=BC=1, AC=\sqrt{2}\). Construct a second \(\triangle DEF\) such that \(\angle E=90^{\circ}, DE=\sqrt{3},EF=1\) and \(DF=2\). Use the triangles to determine the following values:

\[\sin 30^{\circ}, \sin 45^{\circ}, \sin 60^{\circ},\]

\[\cos 30^{\circ}, \cos 45^{\circ}, \cos 60^{\circ},\]

\[\tan 30^{\circ}, \tan 45^{\circ}, \tan 60^{\circ}.\]

\(4\). Construct \(\triangle ABC\) such that \(\angle B=90^{\circ}\) and \(\angle A=\theta\). If \(\tan \theta = \frac{4}{3}\), find the values of \(\sin \theta\) and \(\cos \theta\).

\(5\). Show that, for acute angle \(\theta\), \(\sin^{2}\theta +\cos^{2}\theta = 1\).

\(\textbf{Details and Assumptions:}\)

\(\sin^{2}\theta = (\sin\theta)^{2}, \cos^{2}\theta=(\cos\theta)^{2}\).

In the next note, Introduction to Trigonometry \(2\) (coming soon!), we will define the three basic trigonometric ratios at other angles.

## Comments

Sort by:

TopNewest@Victor Loh Explanatory note well done. – Mardokay Mosazghi · 2 years, 3 months ago

Log in to reply

Superb! – Krishna Ar · 2 years, 3 months ago

Log in to reply

Prove that 1÷seca-1+1÷seca+1=2coseca×cota – Neelesh Sahu · 2 years, 2 months ago

Log in to reply