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 [sin² A - sin² B]

= k [sin (A + B) sin (A – B)]

= k [sin (180ᵒ - 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 =

a² = a² + b² - c²

b² - c² = 0

b² = c²

b = c

Hence, the triangle is isosceles.

5) In ΔABC, prove that: a² = (b + c) ² - 4bc cos².

Solution:

R.H.S. = (b + c) ² - 4bc cos²

= b² + c² + 2bc – 4bc ()

= b² + c² + 2bc – 2bc cos A – 2bc

= b² + c² - 2bc.

= b² + c² - (b² + c² - a²)

= b² + c² - b² - c² + a²

= a² = 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 = 90ᵒ, find sin A and sin B.

Solution:

Ø = =

Ø = =

Ø

Ø sin A = =

Ø 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

Ø a x = b x

Ø a²(b² + c² - a²) = b²(c² + a² - b²)

Ø a²b² + a²c² - a⁴ = b²c² + a²b² - b⁴

Ø a²c² - b²c² - a⁴ = 0

Ø a²(a² - b²) – (a² - b²)(a² + b²) = 0

Ø (a² - b²)(c² - a²- b²) = 0

If a² - b² = 0

Ø a = b

Therefore, ΔABC is isosceles triangle

If c² - a² - b² = 0

Ø c² = a² + b²

Therefore, ΔABC is right triangle.

8) In a ΔABC, if angle B = 60ᵒ, 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) ² - (b) ² = 3ac

a² + c² + 2ac - b² = 3ac

a² + c² - b² = 3ac – 2ac

a² + c² - b² = ac

Ø =

Ø =

Ø cos B =

Ø B = angle 60ᵒ.

9) In ΔABC, prove that : (b² - c²)cot A + (c² - a²)cot B + (a² - b²)cot C = 0

Solution:

L.H.S.

= (b² - c²). = (c² - a²). + (a² - b²).

= (b² - c²). + (c² - a²). + (a² - b²).

= [(b⁴ – c⁴ - a²b² + a²c²)] + (c⁴ - a⁴ - b²c² + a²b²) + (a⁴ - b² - a²c² + b²c²)

= [0]

= R.H.S.

10)In ΔABC, prove that : (a – b)² cos² + (a + b)² sin² = C²

Solution:

L.H.S

= (a – b) ² cos² + (a + b) ² sin²

= (a² + b² - 2ab) () + (a² + b² + 2ab) ()

= - 2ab cos C + a² + b²

= a² + b² - 2ab cos C

= a² + b² - 2ab

= a² + b² - a² - b² + c²

= c²

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

Hence, Proved.