Question

if the sides of a triangle are in a.p and the greatest angle of the triangle is double the smallest angle, the ratio of the sides of the triangle is?

Answer 1

i) As per sine law of a triangle,
"In any triangle, the sides are proportional to sine of the opposite angles".

So applying this here, a/sinA = b/sinB = c/sinC = k ---- (1)

ii) As the sides are in AP, 2b = (a + c)
Substituting from (1), 2k*sin(B) = k(sinA + sinC)
==> 2sin(B) = sin(A) + sin(C)
==> 2sin(B) = 2sin{(A + C)/2}*cos{(A - C)/2} ----- (1)
[Application of sum-product rule identity]

iii) By the given condition, taking So, A + C = 3C and A - C = C and

iv) Substituting all these in (1) above,
sin(3C) = sin(3C/2)cos(C/2)

==> 2sin(3C/2)cos(3C/2) - sin(3C/2)cos(C/2) = 0
[Application of sin(2u) = 2sin(u)cos(u)]

==> sin(3C/2){2cos(3C/2) - cos(C/2)} = 0

So either sin(3C/2) = 0 or 2cos(3C/2) - cos(C/2) = 0
If sin(3C/2) = 0, then C = 0, which is not possible; hence this is rejected.

v) So 2cos(3C/2) - cos(C/2) = 0
==> {cos(3C/2)}/cos(C/2) = 1/2
==> {4cos³(C/2) - 3cos(C/2)}/cos(C/2) = 1/2

==> 4cos²(C/2) - 3 = 1/2
==> cos²(C/2) = 7/8

So cos(C/2) = ±√(7/8)
But in any triangle C/2 is always acute; so cos(C/2) is > 0;
Hence cos(C/2) = √(7/8) only; so cos(C) = 2cos²(C/2) - 1 = 3/4
so sin(C/2) = √{1 - cos²(C/2)} = √(1/8)

vi) Thus of the above, sin(C) = 2sin(C/2)cos(C/2) = 2{√(1/8)}{√(7/8)} = (√7)/4
sin(A) = sin(2C) = 2sin(C)cos(C) = 2{ (√7)/4)(3/4) = 3√7/8
sin(B) = sin(3C) = {3sin(C) - 4sin³(C)} = 3√7/4 - 4(√7/4)³ = 5√7/16

So the proportion of the sides are: a : b : c = sin(A) : sin(B) : sin(C)

= 3√7/8 : 5√7/16 : √7/4 = 6 : 5 : 4

Answer: The sides of the triangle are in proportion as 6 : 5 : 4