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Introduction to Trigonometry Ex- 8.1

 Hi Friends and Champs

Before we proceed to the next exercise let's understand some basic facts about trigonometry.

 Trigonometry is the combination of three Greek words "tri", "gon" and  "metron" . "tri" mean three "gon" means sides and "metron" mean measure. So, trigonometry means to study the relationship between sides and angles of a triangle. 

 

Trigonometry Ratios: 

Let Δ ABC is a right angle triangle , right angle at B, then:

Sin A = BC / CA           i.e. Side opposite to angle A / Hypotenuse
Cos A = AB / CA          i.e. Side adjacent to angle A / Hypotenuse
Tan A = BC / AB          i.e. Side opposite to angle A / Side adjacent to angle A 
Cosec A = CA / BC      i.e Hypotenuse /. Side opposite to angle A
Sec A = CA / AB          i.e. Hypotenuse / Side adjacent to angle A 
Cot A = AB / BC          i.e. Side adjacent to angle A / Side opposite to angle A 

Remember 

Sin θ = 1 / Cos θ            where θ (theta) is the representation of an angle
Cosθ = 1 / Sec θ
Tan θ = 1 / Cot θ

Note : You can easily remember above identities by the rule ( LAL / KKA or लाल / कका in hindi , or PBP / HHB ( पंडित बद्री प्रसाद / हर हर भोले )  in English , use whatever easy to remember for you   

In  PBP / HHB
P -Perpendicular ,  or ( L mean   लंब in Hindi )
B - Base                 or ( A -  mean आधार in Hindi )
H- Hypotenuse       or ( K - mean  in कर्ण Hindi )

in LAL / KKA  or लाल / कका  start reading from left to right and select one letter from up and down in LAL / KKA   like below
Sinθ = L / K , 
Cosθ = A / K , 
Tanθ = L/A
Cosecθ = 1 / Sinθ = K / L
Sec θ = 1 / Cosθ = K / A
Cotθ = 1 / Tanθ = A / L

I hope till now you can easily remember all the above identities, no need for rattafication 😁  .

                                                                        Exercise 8.1

 1. In  Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine : 
 (i) sin A, cos A 
 (ii) sin C, cos C


Solution: 

Since Δ ABC is a right angle , therefore CA2= BC2 + AB2
=>CA2= 72 + 242
=>CA2= 49 + 576 = 625
=>CA= 25

(i)  sin A, cos A

Sin A = BC / CA = 7 / 25 Ans.

Cos A = AB / CA = 24 / 25 Ans.

(ii) sin C, cos C

Sin C = AB / CA = 24 / 25 Ans.

Cos C = BC / CA = 7 / 25 Ans.


 2. In Fig. 8.13, find tan P – cot R.

Fig. 8.13    

Solution:

Since Δ PQR is a right angle at Q, therefore,

PR2= PQ2 + QR2
=>132= 122 + QR2
=>QR2= 169 - 144
=>QR2= 25
=>QR = 5

now tan P - cot R = (5 / 12) - ( 5 / 12 )
                            = 0 Ans.

3. If sin A = 3/4 , calculate cos A and tan A.

Solution:

Since Sin A = Side opposite to angle A / Hypotenuse = 3k / 4k 

where 3 k is the side opposite to angle A and 4k is the Hypotenuse

let draw Δ ABC right angle with same measurement of sides.


then BC = 3k and CA = 4 k 
therefore CA2= BC2 + AB2
=>(4k)2= (3k)2 + AB2
=>AB2= 16k2 - 9k2
=>AB= √7k

Cos A = AB / CA
          =√7k / 4k
          = √7/ 4 Ans
and 
tan A = BC / AB
          =3k / √7k
          = 3 / √7 Ans

4. Given 15 cot A = 8, find sin A and sec A.

solution:

cot A = 8/15 = A/L = Base / Perpendicular

Let the triangle is   

=.> AB = 8 and BC = 15 , so

CA2= BC2 + AB2
=>(CA2= 152 + 82
=>CA2= 225 + 64
=>CA= √289
=>CA = 17

=> SinA = BC / CA = 15/ 17 Ans.
and SecA = CA / AB = 17 / 8 Ans.

5. Given sec θ = 13/12 , calculate all other trigonometric ratios.

since Sec θ = Hypotenuse / Base = 13 /12 
=>let Hypotenuse = 13k
and Base = 12k for  some real number k

Since Hypotenuse2 = Base2 + Perpendicular2
=> 169k2 = 144k2 + Perpendicular2
=> Perpendicular2 = 25k2
=> Perpendicular = 5k
Hence
=> Sinθ = 5/13
=> Cosθ = 12/13
=> Tanθ = 5/12
=> Cosecθ = 13/5
=> Secθ = 13/12
=> Cotθ = 12/5  

 6. If < A and < B are acute angles such that cos A = cos B, then show that < A = < B.

Solution;

Let the triangle is :

Note acute angle mean the angle less than 90 degree.

Since CosA = CosB ( given )
=> CA/ AB  = BC / AB
=> CA = BC 
Since CA  = BC => ABC is an isosceles right angle triangle , so

< A = < B from the property of isosceles triangle.

Also <A + <B + <C = 180° ( property of sum of triangle )

=> < A + <B = 90°  since <C is 90°

since ABC is an   isosceles right angle triangle , therefore

<A = <B = 45°. Ans 

7. If cot θ = 7 / 8 evaluate :
 (i) (1 + sinθ ) (1 - sinθ )  /  (1 + cosθ )(1 - cosθ ) 
(ii) cot2θ

Solution :

Since Cot θ = Base / Perpendicular

let Base is 7K and Perpendicular is 8K for some real number k ≠ zero.
Since Cot θ = Base (B) / Perpendicular (P)
let Base is 7K and Perpendicular is 8K for some real number k ≠ 0.
Therefore the hypotenuse  ( H  )is 
H2 = P2 + B2
=>H2 = 64 + 49
=>H2 = 113
=>H = √113

Hence Sinθ = P/H = 8 / √113
and Cosθ = A/H = 7/√113

i) (1 + sinθ ) (1 - sinθ )  /  (1 + cosθ )(1 - cosθ ) 
= (1 - Sin2θ) / ( 1- Cos2θ)
=(1 - 64/113)/(1-49/113)
=(49/113) /(64/113)
=49 / 64 Ans

(ii) cot2θ
cot2θ = (7/8)2
         = 49/64 Ans

8. If 3 cot A = 4, check whether (1 - tan2A ) / (1+ tan2A)  = cos2A – sin2A or not.

Solution : cot A = 4/3  = B / P

Hence tan A = 1 / cot A = 3/4

and since H2 = P2 + B2
=> H2 = 9 + 16 = 25 
=> H = 5
Hence CosA = 4/5 and SinA = 3/5

LHS = (1 - tan2A ) / (1+ tan2A) = ( 1-9/16) / ( 1 + 9/16) = 7/25
RHS = cos2A – sin2A
         = ( 16/25 - 9/25 )
         = 7/25 = LHS 
        => Given identity is true.

9. In triangle ABC, right-angled at B, if tan A = 1 /√3 
 find the value of:
 (i) sin A cos C + cos A sin C
 (ii) cos A cos C – sin A sin C

Solution:

Since tan A = 1 /√3 = P / B
=.> Let AB = √3k and BC = 1k ,for some real number k ≠ 0.

CA2= BC2 + AB2
=>CA2= 1k2 + √32k2
=>CA2= k2 + 3k2
=>CA2=4k2
=>CA = 2k

=> SinA = BC / CA = 1/ 2
and CosC = BC / CA =1/2
and CosA=AB / CA = √3 / 2
and SinC =AB / CA = √3 / 2

(i) sin A cos C + cos A sin C
    =1/2 x 1/2 + √3 / 2 x √3 / 2
   =1/4 + 3/4
   =1
(ii) cos A cos C – sin A sin C
   = √3 / 2 x 1/2 - 1/2 x √3 / 2
   = 0

10. In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
 
Solution: 
 
PR + QR = 25 ----------------------(1) ( given )

Since
PR2= QR2 + PQ2 (sinceΔ PQR, right-angled at Q )
=> PR2 - QR2 = PQ2
=> ( PR + QR )( PR - QR ) = PQ2
=> 25( PR - QR ) = 25
=> PR - QR = 1----------------(2)

From (1) + (2)
2PR = 26
=> PR = 13
and QR = 12 ( put value or PR in eq(1))

SinP = QR / PR = 12/13 Ans
TanP = QR / PQ = 12 / 5 Ans
CosP = PQ / PR = 5/13 Ans

 11. State whether the following are true or false. Justify your answer.
 (i) The value of tan A is always less than 1.
 (ii) sec A = 12/5  for some value of angle A.
 (iii) cos A is the abbreviation used for the cosecant of angle A. 
 (iv) cot A is the product of cot and A
 (v) sin θ = 4/ 3 for some angle θ.

Solution:


(i) The value of tan A is always less than 1.
Solution:

Let draw an isosceles right-angle triangle, right angle at B.  with sides BC = AB (
then tan A = BC / AB =1
Hence tanA is always less than 1 is not necessary, so the statement is false.

(ii) sec A = 12/5  for some value of angle A.
Solution: 
Since, sec A = H / B =12/5
let H = 12k and B =5k for some real number k≠ 0.

since H2= B2 + P2
=> 144k2= 25k2 + P2
=> 144k2 - 25k2 = P2
=>P2= 119k2
=>P = √119 k , which is also a positive real number , so statement is true.
 
 (iii) cos A is the abbreviation used for the cosecant of angle A. 
Solution:
False , cosecant of angle A is cotA

 (iv) cot A is the product of cot and A
Solution: 
False
cot A mean cosecant of angle A not the product of cot and A
(v) sin θ = 4/ 3 for some angle θ.
Solution:
Since, Sinθ = P/H  =4/3
let P = 4k and H =3k for some real number k≠ 0.

since H2= B2 + P2
=> 9k2= B2 + 16k2
=> 9k2 - 16k2 = B2
=>B2= -7k2
=>B = √-7 k , which is not a positive real number, so statement is False.





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