In a triangle ABC, B = 90 degrees, b + a = 4, then the angle C when area of the triangle is maximum is?
Priyanshu Mohanty , 6 Years ago
Grade 12
Differential Calculus> In a triangle ABC, B = 90 degrees, b + a ...
1 Answers
Arun
Last Activity: 6 Years ago
Dear student
i) The side AB = c
Applying Pythagoras theorem, AB² = AC² - BC² = b² - a² = (b + a)(b - a) = 4(b - a)
= 4(4 - a - a) = 4(4 - 2a)
So c = 2√(4 - 2a)
ii) Area of triangle ABC = (1/2)*(AB)*(BC) [Since angle B = 90 deg]
So, A = (1/2)*2√(4 - 2a)*a = a√(4 - 2a)
iii) For the sake of convenience of differentiating, let us square the above function,
==> A² = a²(4 - 2a) = 4a² - 2a³
Differentiating both sides, 2A*A' = 8a - 6a²
==> A*A' = 4a - 3a²
iv) Equating the above to zero, 4a - 3a² = 0; solving a = 4/3
Again differentiating A*A' = 4a - 3a²
(A')² + A*A" = 4 - 6a
At a = 4/3, A' = 0
So, A*A" = 4 - 8 = -4
==> A" is less than zero at a = 4/3; hence the area is maximum
for the condition a+b = 4
v) For maximum area, a = 4/3 and b = 4 - 4/3 = 8/3
Hence cos(C) = a/b = (4/3)/(8/3) = 1/2
==> C = 60 deg
Answer: Angle C = 60 deg
Applying Pythagoras theorem, AB² = AC² - BC² = b² - a² = (b + a)(b - a) = 4(b - a)
= 4(4 - a - a) = 4(4 - 2a)
So c = 2√(4 - 2a)
ii) Area of triangle ABC = (1/2)*(AB)*(BC) [Since angle B = 90 deg]
So, A = (1/2)*2√(4 - 2a)*a = a√(4 - 2a)
iii) For the sake of convenience of differentiating, let us square the above function,
==> A² = a²(4 - 2a) = 4a² - 2a³
Differentiating both sides, 2A*A' = 8a - 6a²
==> A*A' = 4a - 3a²
iv) Equating the above to zero, 4a - 3a² = 0; solving a = 4/3
Again differentiating A*A' = 4a - 3a²
(A')² + A*A" = 4 - 6a
At a = 4/3, A' = 0
So, A*A" = 4 - 8 = -4
==> A" is less than zero at a = 4/3; hence the area is maximum
for the condition a+b = 4
v) For maximum area, a = 4/3 and b = 4 - 4/3 = 8/3
Hence cos(C) = a/b = (4/3)/(8/3) = 1/2
==> C = 60 deg
Answer: Angle C = 60 deg
Regards
Arun (askIITians forum expert)
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