Simple Applications of Sine and Cosine Formulae

Exercises

1)   Prove that : a sin A b sin B = c sin (A B)

 

Solution:

L.H.S. = asin A bsin B

= ksin A. sin A k sin B. sin B

= k [sin A - sin B]

= k [sin (A + B) sin (A B)]

= k [sin (180 - C) sin (A B)]

= k [sin C sin (A B)]

= Ksin C. sin (A B)

= csin (A B) = R.H.S.

Therefore, L.H.S. = R.H.S.

Hence Proved.

 

2)   In a ΔABC, a = 3, b = 5 and c = 7, find the values of cos A, cos B, cos C.

 

Solution.

cos A = =

= = =

cos B = =

= = =

cos C =

= =

= - = - .

 

 

3)   Prove that : =

Solution.

L.H.S. =

=

=

=

=

= = R.H.S

Therefore, L.H.S. = R.H.S.

Hence, Proved.

4)  In a ΔABC, if cos C =, prove that triangle is isosceles.

Solution:

= cos C

Sin A = 2sin B cos C

ka = 2kb

a =

a = a + b - c

b - c = 0

b = c

b = c

Hence, the triangle is isosceles.

 

5)  In ΔABC, prove that: a = (b + c) - 4bc cos.

Solution:

R.H.S. = (b + c) - 4bc cos

= b + c + 2bc 4bc ()

= b + c + 2bc 2bc cos A 2bc

= b + c - 2bc.

= b + c - (b + c - a)

= b + c - b - c + a

= a = L.H.S.

Therefore, L.H.S. = R.H.S.

Hence, Proved.

 

6)   In ΔABC, if a = 18, b = 24, c = 30 and angle C = 90, find sin A and sin B.

Solution:

     = =

     = =

    

     sin A = =

     sin B = = .

 

7)   If a cos A = b cos B, then the triangle is either isosceles or right angled.

 

Solution.

Given, A cos A = b cos B

     a x = b x

  a(b + c - a) = b(c + a - b)

  ab + ac - a4 = bc + ab - b4

  ac - bc - a4 = 0

  a(a - b) (a - b)(a + b) = 0

  (a - b)(c - a- b) = 0

If a - b = 0

  a = b

Therefore, ΔABC is isosceles triangle

If c - a - b = 0

  c = a + b

Therefore, ΔABC is right triangle.

 

8)   In a ΔABC, if angle B = 60, prove that (a + b + c)(a b + c) = 3ac

 

Solution:

(a + b + c)(a + c b) = 3ac

(a + c + b)(a + c b) = 3ac

(a + c) - (b) = 3ac

a + c + 2ac - b = 3ac

a + c - b = 3ac 2ac

a + c - b = ac

       =

       =

     cos B =

  B = angle 60.

 

9)   In ΔABC, prove that : (b - c)cot A + (c - a)cot B + (a - b)cot C = 0

 

Solution:

L.H.S.

= (b - c). = (c - a). + (a - b).

= (b - c). + (c - a). + (a - b).

= [(b4 c4 - ab + ac)] + (c4 - a4 - bc + ab) + (a4 - b - ac + bc)

= [0]

= R.H.S.

 

10)In ΔABC, prove that : (a b) cos + (a + b) sin = C

 

Solution:

L.H.S

= (a b) cos + (a + b) sin

= (a + b - 2ab) () + (a + b + 2ab) ()

= - 2ab cos C + a + b

= a + b - 2ab cos C

= a + b - 2ab

= a + b - a - b + c

= c

Therefore, L.H.S. = R.H.S.

Hence, Proved.