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CAT 2025 Lesson : Trigonometry & Coordinate - Basics of Triangle

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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 = opphyp\dfrac{\mathrm{opp}}{\mathrm{hyp}} cos θ = adjhyp\dfrac{\mathrm{adj}}{\mathrm{hyp}} tan θ = oppadj=sin⁡ θcos θ\dfrac{\mathrm{opp}}{\mathrm{adj}} = \dfrac{\mathrm{\sin \ θ}}{\mathrm{cos\ θ}}
cosec θ = hypopp\dfrac{\mathrm{hyp}}{\mathrm{opp}} sec θ = hypadj\dfrac{\mathrm{hyp}}{\mathrm{adj}} cot θ = adiopp=cos θsin 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 = 1sin θ\dfrac{1}{\mathrm{sin \ θ}} sec θ = 1cos θ\dfrac{1}{\mathrm{cos \ θ}} cot θ = 1tan 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°0\degree 0 1 0 ∞ 1 ∞
30°30\degree 12\dfrac{1}{2} 32\dfrac{\sqrt{3}}{2} 13\dfrac{1}{\sqrt{3}} 3\sqrt{3} 23\dfrac{2}{\sqrt{3}} 2
45°45\degree 12\dfrac{1}{\sqrt{2}} 12\dfrac{1}{\sqrt{2}} 1 1 2\sqrt{2} 2\sqrt{2}
60°60\degree 32\dfrac{\sqrt{3}}{2} 12\dfrac{1}{2} 3\sqrt{3} 13\dfrac{1}{\sqrt{3}} 2 23\dfrac{2}{\sqrt{3}}
90°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°−60°−90°30\degree-60\degree-90\degree or 45°−45°−90°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°,60°30\degree, 60\degree and 90°90\degree angles of a right-triangle are in the ratio 1:3:21 : \sqrt{3} : 2
2) Sides opposite the 45°,45°45\degree, 45\degree and 90°90\degree angles of a right-triangle are in the ratio 1:1:21 : 1 : \sqrt{2}

Example 1

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

Solution

sin⁡30°cos⁡60°+cot⁡30°cosec⁡60°\sin 30\degree \cos 60\degree + \cot 30\degree \cosec 60\degree

=12×12+3×23=14+2=94\dfrac{1}{2} \times \dfrac{1}{2} + \sqrt{3} \times \dfrac{2}{\sqrt{3}} = \dfrac{1}{4} + 2 = \dfrac{9}{4}

Answer: 94\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°45\degree and 60°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) 3:1\sqrt{3} : 1      (2) (3−1):1(\sqrt{3} - 1) : 1     (3) (2−3):1(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°−45°−90°45\degree - 45\degree - 90\degree triangle where sides should be in the ratio 1:1:21 : 1 : \sqrt{2}. So DB : BC : DC = 1:1:21 : 1 : \sqrt{2}

△ABC is a 30°−60°−90°30\degree - 60\degree - 90\degree triangle where the sides should be in ratio 1:3:21 : \sqrt{3} : 2. So BC : AB : AC = 1:3:21 : \sqrt{3} : 2

Since BC is a common side in both the triangles and the value of BC in both ratios is 1, ABDB=31\dfrac{\mathrm{AB}}{\mathrm{DB}} = \dfrac{\sqrt{3}}{1}

ADDB=AB−DBDB=3−11\dfrac{\mathrm{AD}}{\mathrm{DB}} = \dfrac{\mathrm{AB - DB}}{\mathrm{DB}} = \dfrac{\sqrt{3} - 1}{1}

Answer: (2) (3−1):1(\sqrt{3} - 1) : 1

Example 3

Guru, who is standing on the top of a building, when looking at Partha, who is 1003100\sqrt{3} metres away from the building, notices that the angle of depression is 30°30\degree. As soon as Guru sees Partha, the latter starts running towards the building at 22 m/s. After how many seconds (rounded to 11 decimal place) will the angle of elevation, when Partha sees Guru, be 45°45\degree? [Use 3=1.732\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 = 1003100\sqrt{3} metres.

△ABC is a 30−60−9030 - 60 - 90 triangle where the sides should be in the ratio 1:3:21 : \sqrt{3} : 2. So AC : BC : AB = 1:3:21 : \sqrt{3} : 2

As BC = 1003100\sqrt{3} m, AC = 10033=100\dfrac{100\sqrt{3}}{\sqrt{3}} = 100 m

Let D be the point where Partha sees Guru at an angle of elevation of 45°45\degree, △ADC will be a 45°−45°−90°45\degree - 45\degree - 90\degree triangle, so ratio of AC : DC : AD = 1:1:21 : 1 : \sqrt{2}.

As AC : DC = 1 : 1, DC = 100100 m

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

Time taken = Distancespeed=BC−DC2\dfrac{\mathrm{Distance}}{\mathrm{speed}} = \dfrac{\mathrm{BC - DC}}{2}

= 1003−1002=100(3−1)2=100(1.732−1)2=73.22=36.6\dfrac{100\sqrt{3} - 100}{2} = \dfrac{100(\sqrt{3} - 1)}{2} = \dfrac{100(1.732 - 1)}{2} = \dfrac{73.2}{2} = 36.6s

Answer: 36.636.6 seconds


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