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In a triangle ABC, B = 90 degrees, b + a = 4, then the angle C when area of the triangle is maximum is?

In a triangle ABC, B = 90 degrees, b + a = 4, then the angle C when area of the triangle is maximum is?

Grade:12

1 Answers

Arun
25763 Points
2 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
 
 
Regards
Arun (askIITians forum expert)

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