Question

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=> If cosA+cosB+2cosC=2, where a, b and C are angles of triangle, then AB, BC and CA are in which progression?



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Answer 1

The answer is given below :

Given,

cosA + cosB + 2 cosC = 2

=> cosA + cosB = 2 - 2 cosC

=> cosA + cosB = 2 (1 - cosC)

=> 2 cos{(A + B)/2} cos{(A - B)/2} = 2 [ 1 - {1 - 2 sin²(C/2)}]

=> cos {(π - C)/2} cos{(A - B)/2} = 2 sin²(C/2), since A + B + C = π

=> cos (π/2 - C/2) cos{(A - B)/2} = 2 sin²(C/2)

=> sin(C/2) cos{(A - B)/2} = 2 sin²(C/2)

Now, eliminating sin(C/2), we get

cos{(A - B)/2} = 2 sin(C/2)

Again, multiplying both sides by
[2sin{(A + B)/2}], we get

2 sin{(A + B)/2} cos{(A - B)/2}
= 2 × 2 sin(C/2) sin{(A + B)/2}

=> sinA + sinB = 2 × 2 sin(C/2) cos(C/2),

[since
sin{(A + B)/2}
= sin{(π - C)/2}, since A + B + C = π
= sin(π/2 - C/2)
= cos(C/2)]

=> sinA + sinB = 2 sinC,

[since 2 sin(C/2) cos(C/2) = sinC]

=> a + b = 2c

[Property of a Triangle :

sinA/a = sinB/b = sinC/c,

where a = BC, b = CA and c = AB.]

Therefore, the sides AB, BC and CA of the triangle ABC are in Arithmetic Progression (AP).

Thank you for your question.

Answer 2

⛤Hey There⛤
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Answer :
Arithmetic Progression worked out as under. 

cos A + cos B + 2cos C = 2 
=> cos A + cos B = 2 (1 - cos C) 
=> (b²+c²-a²)/2bc + (c²+a²-b²)/2ca = 2[1 - (a²+b²-c²)/2ab] 
=> a(b²+c²-a²) + b(c²+a²-b²) = 2c[2ab - (a²+b²-c²)] 
=> ab² + ac² -a³ + bc² + ba² - b³ = 2c[2ab - (a²+b²-c²)] 
=> ab² + ba² + ac² + bc² - a³ - b³ = 2c[2ab - (a²+b²-c²)] 
=> ab(a+b) + c²(a+b) - (a+b)(a²-ab+b²) = 2c[2ab - (a²+b²-c²)] 
=> (a+b)(ab+c² - a²+ab-b²) = 2c[2ab - (a²+b²-c²)]
=> (a+b)[2ab - (a²+b²-c²)] = 2c[2ab - (a²+b²-c²)] 
=> a+b = 2c 
=> sides of the triangle are in A.P.
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