How to solve basic problems in trigonometry?(concept-1)

Mathematics tutors give great importance to Trigonometry

Trigonometry is one of the fascinating branches of Mathematics. It deals with the relationships among the sides and angles of a triangle.Word trigonometry was originated from the Greek word, where, ‘TRI‘ means Three‘GON‘ means sides and the ‘METRON’ means to measure. It’s an ancient and probably most widely used branch Mathematics. For basic learning, I am dividing trigonometry in two part:-

1 Trigonometry based on right triangles

1 Trigonometry based on non-right triangles.

In this post, I will only discuss Trigonometry based on right triangles.

In a right triangle, there are three sides hypotenuse (the longest side), adjacent side(base) and the opposite side(perpendicular).

                             <img src="right triangle.png" alt="right triangle">

If we know any two sides of the right triangle, we can find the third side by using Pythagoras theorem:-


In a right triangle, the ratios of sides can be written by many ways like P/H, B/H, P/B, H/P, H/B, B/P. each ratio is called a trigonometric ratio and these are represented by Latin words Sin, Cos, Tan, Cosec, Sec and Cot.
In a right triangle

Sin =P/H, Cos =B/H,  Tan= P/B, Cosec= H/P, Sec =H/B, Cot= B/P,

Mathematics tutors use these three concepts to solve trigonometric problems


In a trigonometric problem, If we are given a trigonometric ratio or asked about it, we will first find all the three sides of the triangle

Using above concept and above formulas, we can solve the first type of trigonometric problems

Example-1  If sin θ = 8/17, find other trigonometric ratios of θ.


Let us draw a ∆ OMP in which ∠M = 90°                                       <img src="right triangle.jpg" alt="right triangle">


Then sin θ = MP/OP = 8/17.

Let MP = 8k and OP = 17k, where k is positive.

By Pythagoras’ theorem, we get

OP2 = OM2 + MP2

⇒ OM2 = OP2 – MP2

⇒ OM2 = [(17k)2 – (8k)2]

⇒ OM2 = [289k2 – 64k2]

⇒ OM2 = 225k2

⇒ OM = √(225k2)

⇒ OM = 15k

Therefore, sin θ = MP/OP = 8k/17k = 8/17

cos θ = OM/OP = 15k/17k = 15/17

tan θ = Sin θ/Cos θ = (8/17 × 17/15) = 8/15

csc θ = 1/sin θ = 17/8

sec θ = 1/cos θ = 17/15 and

cot θ = 1/tan θ = 15/8.

Example-2 If cos θ = 3/5, find the value of (5cosec θ – 4 tan θ)/(sec θ + cot θ)


Let us draw a ∆ ABC in which ∠B = 90°.  <img src="right triangle.png" alt="right triangle">

Let ∠A = θ°

Then cos θ = AB/AC = 3/5.

Let AB = 3k and AC = 5k, where k is positive.

By Pythagoras’ theorem, we get

AC2 = AB2 + BC2

⇒ BC2 = AC2 – AB2

⇒ BC2 = [(5k)2 – (3k)2]

⇒ BC2 = [25k2 – 9k2]

⇒ BC2 = 16k2

⇒ BC = √(16k2)

⇒ BC = 4k

Therefore, sec θ = 1/cos θ = 5/3

tan θ = BC/AB =4k/3k = 4/3

cot θ = 1/tan θ = 3/4 and

csc θ = AC/BC = 5k/4k = 5/4

Now (5cosec θ -4 tan θ)/(sec θ + cot θ)

= (5 × 5/4 – 4 × 4/3)/(5/3 + 3/4)

= (25/4 -16/3)/(5/3 +3/4)

= 11/12 × 12/29

= 11/29

In next post of Mathematics tutors series, we will discuss the remaining two concepts of non-right triangle trigonometry

To get a better understanding of the concept you should check all my posts of Mathematics tutors series. you can check them by clicking on the links given below

First post- concept one (current post)

Second post- concept two(current post)

Third post -Concept three

Fourth post -Concept four

<img src="demo.png" alt="demo">

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