QUESTION BASED ON TRIGONOMETRY

Questions based on trigonometry:

1. In a triangle ABC, right angled at A; if AB= 12cm, AC=5cm. Find all the trigonometric ratios of angle B and C?

  

2. In a triangle ABC right angle that be. If AC equals 5 centimeters, BC equals 3 centimetre.Find 

1. Sin A

2. Cos C

3. Tan A

4. Cosec C

5. Sec^2 A - Tan^2 A

3) In a right-angled triangle. If angle A is acute and cot A= 4/3. Find the remaining trigonometric ratio.

Given: CotA = 4/3 

           Base (AB) = 4

Perpendicular (BC) = 3

By using Pythagoras theorem 

     AC^2 =AB^2+BC^2 

             = 42 + 32

              = 16 +9

      AC^2= 25

      AC = 5

Sin A = Perpendicular/Hypotenuse = 3/5

Cos A = Base/Hypotenuse= 4/5  

Tan A = Perpendicular/Base=3/4     

Cosec A = Hypotenuse/Perpendicular= 5/3

Sec A = Hypotenuse/Base= 5/4

4) For the given figure, if cosˠ=513 and cosβ= 35 Find the length of BD?

In ABC, Cosˠ = 5/13 =BC/AC

                 If BC= 5k, AC= 13k

                 AC^2= BC^2+ BC^2

                 (13k) ^2= 12^2+ (5k) ^2

                  169 k^2 = 144 + 25 k^2

                   169k^2-25k^2 = 144

                    144 k^2= 144

                          k= 1

                   BC =5k

                   BC = 5m

In CDE, Cos β = 3/5 = CD/CE

                 If CD = 3p, CE = 5p

                 CE^2=CD^2 + DE^2

                  (5p) ^2=(3p) ^2 + 8^2

                   25p^2 = 9p^2+ 64

                   25p^2- 9p^2 = 64

                       16p^2 =64

                         p^2=4

                         p=2

                CD= 3P 

                CD = 6m

                BD =BC + CD

                BD = 11m

5) If sin B = ½, Show that 3cosB - 4COS^3B =0

 

Solution : We have, 

                sinB = ½ = perpendicular/hypotenuse

                So let us draw a right triangle ABC, right angled at C such that Perpendicular = AC=1 unit, Hypotenuse = AB = 2 units.

            

Applying Pythagoras theorem in ABC, We obtain

 

AB^2 = BC^2 + AC^2

BC^2 = AB^2 - AC^2

BC^2 = 2^2 - 1^2

BC^2 = 4-1

BC^2 = 3

BC = 3

cos B = BC/ AB = 3 /2

 

3 cosB - 4 cos^2 B = 3x 3 /2- 4 (3 / 2)^3

                               = 33 / 2 - 4x 33 / 8

                               = 33 / 2 - 33 / 2

                                = 0 

Hence proved 

     3 cos B - 4 cos^3 B = 0

Trigonometry

                                                  Trigonometry

INTRODUCTION:

                  Trigonometry is an important branch of mathematics. The word trigonometry is derived from the Greek words 

• Trigonon  means a triangle

•  Mentron  means measure.

Hence trigonometry means the science of measuring triangles, so we can define trigonometry as a study of relationship between the sides and angles of right-angle triangle. in a broader sense it is that branch of mathematics which deals with the measurement of size and the angles of a triangle, and the problems allied with angles.

  

Angle:

         Consider a ray OA, if this ray rotates about its end point O and takes the position of OB then we say that the ∠AOB has been generated. 

          

   An angle is considered as the figure obtained by rotating a given ray about its endpoint.  The revolving ray is called as a generating line of the angles the initial position OA is called the initial side and final position OB is called the terminal side of the angle. The end point about which the ray rotates is called a vertex of the angle. The measure of an angle is the amount of rotation from the initial side to the terminal side.

Notation of angles:

     To indicate an angle any letter of the English alphabet can be used but in trigonometry in general the following Greek letters are used

• Θ (Theta)

•  Φ (Phi)

•  α (Alpha)

•  β (Beta)

•  γ(Gama)

Concept of perpendicular, base and hypotenuse in a right Triangle:

For any acute angle (which is also known as the reference angle) in a right-angled triangle, the side opposite to the acute angle is called the perpendicular; the side adjacent to it is called the base and the side opposite to the right angle is called the hypotenuse.

Trigonometric ratios

                The most important task of trigonometry is to find the remaining sides and angles of a triangle when some of its sides and angles are given. This problem is solved by using some ratios of sides of a triangle with respect to its acute angle. The ratio between the length of a pair of two sides of a right-angled triangle is called trigonometric ratio. 

                  The three sides of a right-angle triangle give 6 trigonometric ratios namely sine, cosine, tangent, cotangent, secant and cosecant. In short, these ratios are written as sin, cos, tan, cot, sec, cosec.

In a right-angled triangle ABC for acute angle A:

1. Sine (sin) is defined as the ratio between the length of the perpendicular and hypotenuse. 

      Sin A= (Perpendicular / Hypotenuse) = (BC / AC)

2. Cosine (cos) Is defined as the ratio between the length of base and hypotenuse.

       Cos A = (Base / hypotenuse) = (AB / AC)

3. Tangent (tan) Is defined as the ratio between the length of perpendicular and base 

         Tan A = (Perpendicular / base) = (BC / AB)

4. Cotangent (cot) Is defined as the ratio between the length of base and perpendicular

          Cot A = (Base / perpendicular) = (AB / BC)

5. Secant (sec) Is defined as the ratio between the lengths of hypotenuse and base.

        Sec A = (Hypotenuse / base) = (AC / AB)

6. Cosecant (cosec) Is defined as the ratio between the length of hypotenuse and perpendicular.

                  Cosec A = (hypotenuse / perpendicular) = (AC / BC)

Similarly for acute angle C In their right-angle triangle ABC,

you can refer to the image given above. 

1. Sin C= Perpendicular/Hypotenuse = AB/AC

2. Cos C= Base/Hypotenuse = BC/AC

3. Tan C= Perpendicular/Base = AB/BC

4. Cot C = Base/Perpendicular = BC/AB

5. Secant C = Hypotenuse/Base = AC/BC 

6. Cosec C = Hypotenuse/perpendicular = AC/AB

Theorem: The trigonometric ratios are the same for the same angle. 

Proof: Let ∠XAY = Be an acute angle, with the initial side AX and

the terminal side AY. Let P and Q two points, both different from A on

the terminal AY. Draw perpendiculars PM and QN from

P & Q respectively on AX. We have to prove that the

trigonometric ratio of angle are same in both

the triangles AMP and ANQ.

In AMP and ANQ we have, ∠MAP = ∠XAY = ∠NAQ and ∠AMP = ∠ANQ = One right angle.

Thus, the corresponding angles of the triangles AMP and ANQ

are equal and, therefore by AAA similarity criterion we obtain. 

     ΔMAP~NAQ ⇒ AP/AQ=PM/QN=AM/AN ⇒ PM/AP=QN/AQ⇒ 1

In ΔAMP, we have

       sinθ = PM/AP sinθ = QN/AQ (using 1)

This shows that the value of sinθ is dependent on the position of the point P. Similarly, it can be proved that the other trigonometric ratios are independent of the position of the point P. 

Points to remember:

• Trigonometry is defined as the study of the relationship

between the sides and angles of a right-angle triangle. 

• Trigonometric ratios are defined for an acute angle

• Sin A= Perpendicular/Hypotenuse = BC/AC

• Cos A = Base/ hypotenuse = AB/AC

• Tan A = Perpendicular/ base = BC/AB

•  Cot A = Base/ perpendicular = AB/BC

• Sec A = Hypotenuse/ base = AC/AB

• Cosec A = hypotenuse/ perpendicular = AC/BC

• Cosec A = 1/Sin A

• Sec A = 1/Cos A

• Cot A = 1/Sec A