Question

In a triangle ABC, angleC = 3 angel B = 2(angle A + angleB). Find all the angles in degrees.

Answer 1

❮SOLUTION❯

Let ∠A = x and ∠B = y

Given ∠C = 3 ∠B =2 (∠A + ∠B)

⇒∠C = 3 ∠B and 3 ∠B = 2(∠A + ∠B)

⇒ ∠C = 3y and 3y = 2(x + y)

⇒∠C = 3y and 2x - y = 0 .....(i)

In ∆ABC, ∠A + ∠B + ∠C = 180

⇒ x + y + 3y = 180° ⇒ x + 4y = 180° .....(ii)

Multiplying equation (i) by 4, we get

8x - 4y = 0.....(iii)

On adding equation (ii) and (iii), we get

9x = 180° => x = 20°

substituting this value of x in eq. (i), we get

(2 × 20)° - y = 0 ⇒y = 40°

∴ ∠C = 3y = (3 × 40)° = 120°

Hence, ∠A = 20°, ∠B = 40° ,∠C = 120°

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Answer 2

Answer:

Step-by-step explanation:

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