Question

In Δ ABC,angle C =3angle B=2(angle A + angle B),then angle B=? Find?

Answer 1

Let ∠ A = x

∠ B = y

As mentioned in the question ∠ C = 3 ∠ B = 3y

∠ C is also given as 2(∠ A + ∠ B)

= 2(x+y)

So,

3y = 2(x+y)

y = 2x

So ∠ C = 3y = 6x

We know that sum of all the 3 ∠s in a tri∠ = 180°

Therefore,

∠ A+ ∠ B + ∠ C = 180

x+ 2x+ 6x = 180

x = 20

Hence ∠ A = 20

∠ B = 40

∠ C = 120

There’s an alternate method too:

let <C=3<B=2<(A+B)=x

 

<C=x, -----(1)

3<B=x

<B=x/3----(2)

2<(A+B)=x

<A+<B=x/2

<A=x/2 -<B

<A=x/2-x/3-----(3)

sum of the three angles in a triangle is 180degree

<A+<B+<C=180

x/2-x/3+x/3+x=180[from(1),(2),(3)]

(3x-2x+2x+6x)/6=180

9x/6=180

x=180*6/9

x=20*6

x=120

 

therefore

A= x/2-x/3=120/2-120/3=60-40=20

B=x/3=120/3=40

C=x=120

 

Or,

C = 3B 
C = 2(A + B) = 2A + 2B 
therefore, 3B = 2A + 2B 
B = 2A and C = 3B = 6A 

in terms of A, the three angles are A, 2A, and 6A 
9A = 180, A = 20 

A = 20, B = 40, C = 120

check: C = 3B ==> 120 = 3(40)  
            C= 2(A + B) ==> 120 = 2(20+ 40)

Hope This Helps :)