1. Introduction

The first part of this lesson deals with trigonometry. Trigonometry means “Measure of Triangle” where Trigonon means Triangle and Metron means to measure. The basis for trigonometry comes from the relationship between the sides and angles of right-angled triangles. Trigonometry is used in many fields of science (like astronomy, geography, engineering, meteorology) for finding distances between two objects or points in observation.

The second part of this lesson deals with Cartesian coordinate geometry in the 2-dimensional plane. This will include concepts such as distance between points, angles and intersection of lines, curves, etc.

2. Trigonometry

Angles are commonly measured in 2 ways – Degrees and Radians. The relationship between these two, i.e. π rad = 180$\degree$ has been discussed and explained in the Lines & Triangles Lesson.


Where θ is an angle opposite one of the non-hypotenuse sides, i.e. an acute angle, there are three possible sides with respect to this angle – hypotenuse of the triangle (hyp), side opposite the angle (opp) and the non-hypotenuse side that is adjancent the angle (adj). The ratios of the sides of a right angled triangle are called trigonometric ratios. The six possible trigonometric ratios are


sin θ = $\dfrac{\mathrm{opp}}{\mathrm{hyp}}$	cos θ = $\dfrac{\mathrm{adj}}{\mathrm{hyp}}$	tan θ = $\dfrac{\mathrm{opp}}{\mathrm{adj}} = \dfrac{\mathrm{\sin \ θ}}{\mathrm{cos\ θ}}$
cosec θ = $\dfrac{\mathrm{hyp}}{\mathrm{opp}}$	sec θ = $\dfrac{\mathrm{hyp}}{\mathrm{adj}}$	cot θ = $\dfrac{\mathrm{adi}}{\mathrm{opp}} = \dfrac{\mathrm{cos \ θ}}{\mathrm{sin \ θ}}$

As the ratios in the second row are the reciprocal of the ratios above,


cosec θ = $\dfrac{1}{\mathrm{sin \ θ}}$	sec θ = $\dfrac{1}{\mathrm{cos \ θ}}$	cot θ = $\dfrac{1}{\mathrm{tan \ θ}}$

2.1 Common Trigonometric Values

The following table provides the values of Trigonometric ratios for commonly used angles.

	Sin	Cos	Tan	Cot	Sec	Cosec
$0\degree$	0	1	0	∞	1	∞
$30\degree$	$\dfrac{1}{2}$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{\sqrt{3}}$	$\sqrt{3}$	$\dfrac{2}{\sqrt{3}}$	2
$45\degree$	$\dfrac{1}{\sqrt{2}}$	$\dfrac{1}{\sqrt{2}}$	1	1	$\sqrt{2}$	$\sqrt{2}$
$60\degree$	$\dfrac{\sqrt{3}}{2}$	$\dfrac{1}{2}$	$\sqrt{3}$	$\dfrac{1}{\sqrt{3}}$	2	$\dfrac{2}{\sqrt{3}}$
$90\degree$	1	0	∞	0	∞	1

Note: It is not required to memorise the entire table. While, the sine values should suffice as explained in the videos, we recommend that you memorise the Sin and Tan Values listed above for speed in exams.

Right-angled triangles with the above mentioned angles can be either $30\degree-60\degree-90\degree$ or $45\degree-45\degree-90\degree$ triangles. While we have discussed this in the Lines & Triangles lesson, we shall look at the same in a couple of examples below. Note the following:

1) Sides opposite the $30\degree, 60\degree$ and $90\degree$ angles of a right-triangle are in the ratio $1 : \sqrt{3} : 2$
2) Sides opposite the $45\degree, 45\degree$ and $90\degree$ angles of a right-triangle are in the ratio $1 : 1 : \sqrt{2}$

Example 1

What is the value of sin $30\degree$ cos $60\degree$ + cot $30\degree$ cosec $60\degree$?

Solution

$\sin 30\degree \cos 60\degree + \cot 30\degree \cosec 60\degree$

=$\dfrac{1}{2} \times \dfrac{1}{2} + \sqrt{3} \times \dfrac{2}{\sqrt{3}} = \dfrac{1}{4} + 2 = \dfrac{9}{4}$

Answer: $\dfrac{9}{4}$

Example 2

A flagpost is on top of a building. The angles of elevation for a man lying on the ground while looking at the base and the top of the flagpost are $45\degree$ and $60\degree$ respectively. What is the ratio of the height of the flagpost to that of the building? (Assume that the man's vision is from ground level)

(1) $\sqrt{3} : 1$ (2) $(\sqrt{3} - 1) : 1$ (3) $(2 - \sqrt{3}) : 1$ (4) None of the above

Solution

Let AD and DB be the height of the flagpost and the building respectively.

△DBC is a $45\degree - 45\degree - 90\degree$ triangle where sides should be in the ratio $1 : 1 : \sqrt{2}$. So DB : BC : DC = $1 : 1 : \sqrt{2}$

△ABC is a $30\degree - 60\degree - 90\degree$ triangle where the sides should be in ratio $1 : \sqrt{3} : 2$. So BC : AB : AC = $1 : \sqrt{3} : 2$

Since BC is a common side in both the triangles and the value of BC in both ratios is 1, $\dfrac{\mathrm{AB}}{\mathrm{DB}} = \dfrac{\sqrt{3}}{1}$

$\dfrac{\mathrm{AD}}{\mathrm{DB}} = \dfrac{\mathrm{AB - DB}}{\mathrm{DB}} = \dfrac{\sqrt{3} - 1}{1}$

Answer: (2) $(\sqrt{3} - 1) : 1$

Example 3

Guru, who is standing on the top of a building, when looking at Partha, who is $100\sqrt{3}$ metres away from the building, notices that the angle of depression is $30\degree$. As soon as Guru sees Partha, the latter starts running towards the building at $2$ m/s. After how many seconds (rounded to $1$ decimal place) will the angle of elevation, when Partha sees Guru, be $45\degree$? [Use $\sqrt{3} = 1.732$ and assume that the heights of Guru and Partha are equal]

Solution

Let A and B be the points where Guru and Partha are standing respectively. The distance between Partha and the building is BC = $100\sqrt{3}$ metres.

△ABC is a $30 - 60 - 90$ triangle where the sides should be in the ratio $1 : \sqrt{3} : 2$. So AC : BC : AB = $1 : \sqrt{3} : 2$

As BC = $100\sqrt{3}$ m, AC = $\dfrac{100\sqrt{3}}{\sqrt{3}} = 100$ m

Let D be the point where Partha sees Guru at an angle of elevation of $45\degree$, △ADC will be a $45\degree - 45\degree - 90\degree$ triangle, so ratio of AC : DC : AD = $1 : 1 : \sqrt{2}$.

As AC : DC = 1 : 1, DC = $100$ m

So, the time taken by Partha to see Guru at an angle of elevation of $45\degree$,

Time taken = $\dfrac{\mathrm{Distance}}{\mathrm{speed}} = \dfrac{\mathrm{BC - DC}}{2}$

= $\dfrac{100\sqrt{3} - 100}{2} = \dfrac{100(\sqrt{3} - 1)}{2} = \dfrac{100(1.732 - 1)}{2} = \dfrac{73.2}{2} = 36.6$s

Answer: $36.6$ seconds