If cos A = 12/13 , cosB = 3/5 , cos C = 63 / 65 and A, B, C are positive acute angles,

Question

then value of A + B + C = ?

The options are :- 

a) pi 

b) pi / 2 

c) 2 pi / 3 

d) none of these

Answer 1

cos A = 12/13, cos B = 3/5, and cos C = 63/65.

If cos A = 12/13, then sin A = √(1 - cos² A) = √(1 - (12/13)²) = √(1 - (144/169)) = √(169 - 144)/169 = √(25/169) = 5/13.

If cos B = 3/5, then sin B = √(1 - cos² B) = √(1 - (3/5)²) = √(1 - (9/25)) = √(25 - 9)/25 = √(16/25) = 4/5.

If cos C = 63/65, then sin C = √(1 - cos² C) = √(1 - (63/65)²) = √(1 - (3969/4225)) = √( 4225 - 3969)/4225 = √(256/4225) = 16/65.

cos (A + B + C) = cos((A + B) + C)

Apply formulas : cos(A + B) = cos A cos B - sin A sin B,

                              sin(A + B) = sin A cos B + cos A sin B.

= cos(A + B) cos C - sin(A + B) sin C

= [(cos A cos B - sin A sin B)cos C] - [(sin A cos B + cos A sin B)sin C]

= cos A cos B cos C - sin A sin B cos C - sin A cos B sin C - cos A sin B sin C

Substitute the values cos A = 12/13, cos B = 3/5, cos C = 63/65, sin A = 5/13, sin B = 4/5, and sin C = 16/65.

= (12/13)(3/5)(63/65) - (5/13)(4/5)(63/65) - (5/13)(3/5)(16/65) - (12/13)(4/5)(16/65)

= (2268 - 1260 - 240 - 768)/4225

= (2268 - 2268)/4225

= 0/4225

= 0.

cos (A + B + C) = 0

A + B + C = cos^{- 1} [cos (π/2)]

∴ A + B + C = π/2.

 

Option b is the correct choice.