Chapter 8. Introduction to Trigonometry

Class 10 Maths Chapter 8. Introduction to Trigonometry Exercise 8.1 Solutions

1. In , right-angled at B ,  . Determine:   (i)  sinA  , cosA   (ii)  sinC , cosC

Solution :  Here , AB=24 cm , BC=7 cm

In∆ABC, we have

 

 

 

 

 

(i)    and

(ii)    and

2. In Fig. 8.13, find  . 

Solution : Here, PQ=12 cm  , PR=13 cm  

In∆PQR , we have

    

 

 

 

 

    

 Now ,  

and

3. If   , calculate cosA and tanA .

Solution :  Given 

let

 

 

In∆ABC , we have

  

 

 

 

and

4. Given  , find sinA and secA .

Solution :  Given,

Let, 

 

 In∆ABC , we have

  

 

 

 

 

and

5. Given  , calculate all other trigonometric ratios .

Solution :  Given,

Let  

 

In∆ABC , we have

  

 

 

 

 

and

6. If  and  are acute angles such that , then show that  .

Solution :  Given, 

      

So, ABC is an isosceles triangle .

          proved .

7. If  , evaluate : (i)         (ii)

Solution: Given,

Let, 

 

In∆ABC , we have

 

 

 

 

and

(i) We have,

(ii) We have,

 8. If  , check whether  or not .

Solution: Given ,

Let, 

 

In∆ABC , we have

 

 

 

 

 

LHS:  

RHS :

  

9. In triangle ABC, right-angled at B , if   , find the value of : (i)  (ii)  

Solution : Given, 

Let 

 

In∆ABC , we have

 

 

 

 

 

(i) We have,

(ii) We have,

10. In , right-angled at Q,  and . Determine the values of and  .

Solution :  Here , 

  and   

   In  , we have

 

 

 

 

 

 

    

     

 

11. State whether the following are true or false . Justify your answers  :

(i) The value of tanA is always less than 1 .

(ii)   for some value of angle A.

(iii)  cosA is the abbreviation used for the cosecant of angle A .

(iv)  cotA is the product of cot and A .

(v)    for some angles θ .

Solution:  (i)  False ,because the value of tanA is not always less than 1. For example : .

(ii) True , because the hypotenuse is the longest side in a right triangle,

     then the value of secA is always greater than or equal to 1 .

(iii) False, because cosA is the abbreviation used for the cosine of angle A .

(iv) False , because cotA is the ratio of base and perpendicular of the right triangle .

(v) False , because the hypotenuse is the longest side in a right triangle, then the value of sinθ is always less than 1 .

 Class 10 Maths Chapter 8. Introduction to Trigonometry Exercise 8.2 Solutions

1. Evaluate the following : (i)  (ii)  (iii)       (iv)     (v)  

Solution :  (i) We have,

 

(ii)  We have,

(iii) We have,  

(iv)  We have,

 (v) We have, 

2. Choose the correct option and justify your choice  : (i)  

 (A)             (B)        (C)      (D)

Solution: (A)            

[Hint:  We have,

]    

(ii)

(A)    (B)  1      (C)     (D)  0

Solution: (D)  0

[ Hint:  We have,:

]

(iii)  is true when  

  (A) 0°          (B) 30°         (C) 45°          (D) 60°

Solution : (A)  0°

[ Hint: Putting 

LHS : 

RHS :   ]

(iv) 

(A)    (B)    (C)   (D)

Solution : (C)  

[ Hint: We have,

]

3. If   and ,find and .

Solution :  We have,

 

and

 

  

Putting  in equation  , we have

      

 

   

Therefore , the value of A and B are 45° and 15° respectively .

4. State whether the following are true or false . Justify your answer .

(i)  sin(A+B) = sinA+sinB

 (ii) The value of sinθ increases as θ increases .

(iii) The value of cosθ increases as θ increases .

(iv)  sinθ = cosθ for all values of θ .

(v)  cotA is not defined for A = 0°

Solution :  (i) sin(A + B) = sinA + sinB

False , Because if you are putting the value of A and B then both sides are not equal .

 (ii) The value of sinθ increases as θ increases .

(ii)  True , because the value of sinθ increases asθ increases .

(iii) False , because the value of cosθ increases as θ increases .

      [ i.e., the value of cosθ increases as θ decreases .]

(iv) False , only for θ=45° is equal and other value of θ is not equal both sides .

(v) True, because the value of  A=0° is not defined for cotA .

Class 10 Maths Chapter 8. Introduction to Trigonometry Exercise 8.3 Solutions

1. Evaluate : (i)    (ii)    (iii) (iv)

Solution : (i) We have,

(ii) We have ,

(iii) We have,

(iv) We have,

2. Show that :

(i)    

(ii)   

Solution :

(i) LHS : 

 

 

 

    RHS

(ii)   LHS : 

 

 

 

    RHS  

3. If   where  is an acute angle , find the value of A .

Solution: We have,

 

Since 90° – 2A  and A – 18° are both acute  angles .

     

   

 

 

Therefore, the value A is 36° .

4. If tanA = cotB , prove that  .

Solution:  Given,  

   

Since A and  are both acute angles .

  

  Proved .

5. If  where 4A is an acute angle, find the value of A .

Solution : We have,

 

Since  and  are both acute  angles .

  

  

     

Therefore, the value A is 22° .

6. If A , B and C are interior angles of a triangle ABC , then show that .

Solution :  Since, A , B and C are interior angles of a triangle ABC respectively .

7. Express  in terms of trigonometric ratios of angles between 0° and 45° .

Solution :  We have ,

 

 

Class 10 Maths Chapter 8. Introduction to Trigonometry Exercise 8.4 Solutions

1. Express the trigonometric ratios sinA , secA and tanA in terms of cotA  .

Solution : We know that ,

 

and

2. Write all the other trigonometric ratios of in terms of secA .

Solution : We have ,

and 

3. Evaluate : (i)        (ii)

Solution: (i) We have ,

             

(ii) We have, 

 

 

 

 

4. Choose the correct option . Justify your choice .

(i)  

 (A)  1        (B) 9        (C) 8        (D)  0

Solution: (B) 9

[ hints : 

  ]

(ii)

   (A) 0       (B)  1      (C)   2       (D) 

Solution:  (C)  2

[Hints:  We have ,

  ]

(iii)   

     (A)      (B)      (C)      (D)      

Solution:  (D) cosA

[ Hints: We have,

  ]  

(iv)   

(A)        (B)         (C)        (D)

Solution: (D)  

[ Hints:  We have ,

  ]

5. Prove the following identities , where the angles involved are acute angles for which the expressions are defined .

 (i)   

Solution : (i)   L.H.S :

  =  R.H.S   Proved.

(ii)  

Solution: L.H.S :

= R.H.S.    Proved.

(iii)  

 [Hint : Write the expression in terms of sinθ and cosθ]

Solution :  L.H.S : 

= R.H.S.  Proved

(iv)  

 [Hint : Simplify LHS and RHS separately]

Solution:  L.H.S :

RHS : 

LHS = RHS    Proved

(v)  , using the identity  .

Solution: LHS : 

   RHS   Proved.

(vi) 

Solution : LHS : 

  = RHS Proved

(vii) 

Solution : L.H.S. :

= R.H.S.  Proved .

(viii)    

Solution :  L.H.S.  :

 

       R.H.S     Proved.

(ix) 

 [ Hint :Simplify LHS and RHS separately ]

Solution: LHS :

RHS : 

LHS = RHS   Proved .

(x) 

Solution:   First part : 

 Second part : 

      First part = Second part = Third part .  Proved.