Question

In triangle ABC tanA:tanB:tanC =1:2:3 then sinA:sinB:sinC =

Answer 1

Answer:

√5 : 2√2: 3

Step-by-step explanation:

TanA : TanB : TanC =1 : 2 : 3

Let TanA = x, TanB = 2x and TanC = 3x.

Since A,B,C are angles of a triangle, sum of the angles = 180

A+B+C = 180

A+B = 180 - C

Tan(A+B) = Tan(180-C)

TanA+TanB/ 1 - TanATanB = - TanC.

=> TanA + TanB = - TanC+ TanATanBTanC

=> TanA+TanB+TanC = TanATanBTanC

Substituting the values

=>  x + 2x + 3x = x * 2x * 3x

=> 6x = 6x³

=> x² = 1

=> x = 1 ( please note x = -1 is not considered as it will give an obtuse angle)

Thus TanA = 1, TanB = 2, TanC = 3.

Now

SinA = TanA/SecA = TanA/√ (1 + Tan²A) = 1 /√2

SinB = TanB/SecB = TanB/√ (1 + Tan²B) = 2/√5

SinC = TanC/SecC = TanC/√ (1 + Tan²C) = 3/√10

So SinA:SinB:SinC = 1/√2 : 2/√5 : 3/√10

                               = √5/√10 : 2√2/√10 : 3/√10

                               = √5 : 2√2: 3