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If in a triangle ABC, acos2 (c/2) + ccos2 (A/2) = 3b/2, then its sides a, b, c are in
- a)A. P.
- b)G.P.
- c)H.P.
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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If in a triangle ABC, acos2 (c/2) + ccos2 (A/2) = 3b/2,then its sides ...
Given: acos(2c/2)ccos(2A/2) = 3b/2
To find: The relation between sides a, b, and c
Solution:
We know that cos(2x) = 1 - 2sin^2(x) and cos^2(x/2) = (1 + cos(x))/2
Substituting the above values in the given equation, we get:
a(1 - sin^2(c))c(1 + cos(A))/2 = 3b/2
=> a(1 - sin(c))(1 + cos(A))/2 = 3b/2c
=> a/2(1 - sin(c))(1 + cos(A)) = 3b/4c
=> a/2(1 - sin(c)) + a/2 cos(A)(1 - sin(c)) = 3b/4c
=> a/2(1 - sin(c)) + b/2 sin(A)(1 - sin(c)) = 3b/4c
=> (a + b/2(2sin(A/2)cos(A/2)))(1 - sin(c)) = 3b/4c
=> (a + b sin(A))(1 - sin(c)) = 3b/4c
=> a(1 - sin(c)) + b sin(A)(1 - sin(c)) = 3b/4c
=> a - a sin(c) + b sin(A) - b sin(A) sin(c) = 3b/4c
=> a + b sin(A) - a sin(c) - b sin(A) sin(c) = 3b/4c
=> a + b sin(A) - b sin(A) sin(c) = a sin(c) + 3b/4c
=> a(1 - sin(c)) = b(3/4 - sin(A) sin(c))
=> a = b(3/4 - sin(A) sin(c))/(1 - sin(c))
=> a = b(3/4 + sin(B) sin(c))/(1 - sin(c)) [Using sin(A) = sin(B) and sin(B) = sin(c)/2R]
=> a = b(3/4 + (a/2R)(b/2R))/(1 - (a/2R))
=> a(1 - (a/2R)) = b(3/4 + (a/2R)(b/2R))
=> a^2/2R = b(3/4) + ab^2/4R^2 - b/2
=> a^2 = b^2(3/2R) + ab^2/2R^2 - bR
=> a^2 = (3b^2 + 2bc)R/2R^2
=> a^2 = (3b^2 + 2bc)/(2R)
=> a = b(3/2sin(A/2))/sin(B/2)
Therefore, a, b, and c are in A.P.
Hence, option A is the correct answer.
To find: The relation between sides a, b, and c
Solution:
We know that cos(2x) = 1 - 2sin^2(x) and cos^2(x/2) = (1 + cos(x))/2
Substituting the above values in the given equation, we get:
a(1 - sin^2(c))c(1 + cos(A))/2 = 3b/2
=> a(1 - sin(c))(1 + cos(A))/2 = 3b/2c
=> a/2(1 - sin(c))(1 + cos(A)) = 3b/4c
=> a/2(1 - sin(c)) + a/2 cos(A)(1 - sin(c)) = 3b/4c
=> a/2(1 - sin(c)) + b/2 sin(A)(1 - sin(c)) = 3b/4c
=> (a + b/2(2sin(A/2)cos(A/2)))(1 - sin(c)) = 3b/4c
=> (a + b sin(A))(1 - sin(c)) = 3b/4c
=> a(1 - sin(c)) + b sin(A)(1 - sin(c)) = 3b/4c
=> a - a sin(c) + b sin(A) - b sin(A) sin(c) = 3b/4c
=> a + b sin(A) - a sin(c) - b sin(A) sin(c) = 3b/4c
=> a + b sin(A) - b sin(A) sin(c) = a sin(c) + 3b/4c
=> a(1 - sin(c)) = b(3/4 - sin(A) sin(c))
=> a = b(3/4 - sin(A) sin(c))/(1 - sin(c))
=> a = b(3/4 + sin(B) sin(c))/(1 - sin(c)) [Using sin(A) = sin(B) and sin(B) = sin(c)/2R]
=> a = b(3/4 + (a/2R)(b/2R))/(1 - (a/2R))
=> a(1 - (a/2R)) = b(3/4 + (a/2R)(b/2R))
=> a^2/2R = b(3/4) + ab^2/4R^2 - b/2
=> a^2 = b^2(3/2R) + ab^2/2R^2 - bR
=> a^2 = (3b^2 + 2bc)R/2R^2
=> a^2 = (3b^2 + 2bc)/(2R)
=> a = b(3/2sin(A/2))/sin(B/2)
Therefore, a, b, and c are in A.P.
Hence, option A is the correct answer.
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If in a triangle ABC, acos2 (c/2) + ccos2 (A/2) = 3b/2,then its sides ...
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If in a triangle ABC, acos2 (c/2) + ccos2 (A/2) = 3b/2,then its sides a, b, c are ina)A. P.b)G.P.c)H.P.d)none of theseCorrect answer is option 'A'. Can you explain this answer?
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