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In Triangle ABC, CosA /a = CosB / b = Cos C / c and the side a = 2. find the area of the triangle.

In Triangle ABC,
CosA /a = CosB / b = Cos C / c
and the side a = 2. find the area of the triangle.

Grade:9

5 Answers

Ashwin Muralidharan IIT Madras
290 Points
9 years ago
Hi Anushka,
In any triangle, we have
a/SinA = b/SinB = c/SinC (this is the Sin rule applicable for any triangle).
And here CosA/a = CosB/b = CosC/c
So CosA/SinA = CosB/SinB = CosC/SinC
or CotA = CotB = CotC
In a triangle this is possible only when A = B = C = 60 degrees.
So the trianle is equilateral.
And hence a = b = c = 2
Or area = [root(3)/4] * (2^2) = root(3). { Area of an equilateral triangle is [root(3)/4]*(a^2) }

Hope that helps.
Best Regards,
Ashwin (IIT Madras).
Swapnil Saxena
102 Points
9 years ago
Using th SIne law
Sin A /a = Sin B / b = Sin C / c  ------(1)
CosA /a = CosB / b = Cos C / c ------(2)
Solving Both the the Equations we get
tan A =tan B= tan C
This can Be possible only when  A = B = C =60
This means the triangle is equilateral.
thus the area is ((3)1/2 /4)(a)2 = ((3)1/2 /4)(2)2= (3)1/2
Surya Pasricha
36 Points
5 years ago
this implies 
A=B=C=60 degrees
so the area =under root ¾ into side square=
under root ¾ into 4=under root 3
Nihal Shah
24 Points
4 years ago
In ∆ABC , THE sides b and angle B are fixed and remaining elements undergo small changes . Show that A∆c + cos C ∆a = 0
Kushagra Madhukar
askIITians Faculty 629 Points
9 months ago
Dear student,
Please find the solution to your problem below.
 
Using the Sine law
SinA /a = SinB / b = SinC / c  ------(1)
CosA /a = CosB / b = CosC / c ------(2)
Solving Both the the Equations we get
tanA = tanB = tanC
This can be possible only when  A = B = C = 60
This means the triangle is equilateral.
Hence, a = b = c = 2
Thus the area is √3/4 x a2 = √3/4 x 22 = √3
Hence, the area of triangle is √3
 
Hope it helps.
Thanks and regards,
Kushagra
 

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