Question

in triangle abc if a=2b and A=3B. then find the value of c/b. a, b,c are sides opp to angle A,B,C respectively.​

Answer 1

Answer:

Explanation:

Let ABC be a triangle with A = 3B and a = 2b. Using the law of sines, we take:

a/(sinA) = b/(sinB) => 2b/(sin(3B)) = b/(sinB) => 2sinB = sin(3B) =>

2sinB = 3(sinB)(cosB)^2 - (sinB)^3 => 2 = 3(cosB)^2 - (sinB)^2 =>

2 = 3(cosB)^2 + (cosB)^2 - 1 => 4(cosB)^2 = 3 => (cosB)^2 = 3/4 =>

cosB = sqr(3/4) => cosB = (sqr(3))/2 => B = 30 degrees.

Therefore, the triangle BAC is right triangle with A = 90.

Now, by using some trigonometry and Pythagoras’ theorem we try to evaluate a:

c/(2b) = cos(30) => c = 2b(cos(30)) => c = 2b[(sqr(3))/2] => c = b(sqr(3))

Therefore, From a = 2b, we take:

[b(sqr(3))]^2 + b^2 = (2b)^2 => 3(b^2) + b^2 = (2b)^2 => 4b^2 = 4b^2

This means that we do not have enough information to find the value of a. We need the value of b or the value of c to be given initially.