Question

In a triangle abc angle c=3,angle b=2(angle a+angle b). find the three angles ?

Answer 1

First of all, the correct ques is

angC= 3angB = 2(angA+angB)

ang C = 2 (ang A+angB)

we know in triangle ABC

A+B+C=180•

A+B+2(A+B)=180•

3(A+B)=180•

A+B=60•.....eqn 1

Now we r given

3B=2(A+B)

3B=2A+2B

B=2A...eqn2

putting this in eqn 1

A+2A=60•

3A=60•

A=20•

From eqn 2

B=2*20

B=40•

A+B+C=180

20+40+C= 180

60+C=180

C=120•

Answer 2

Answer:

Step-by-step explanation:

Solution :-

Let ∠A = x° and ∠B = y°

Then,

∠C = 3∠B = (3y°)

Now,

∠C = 2(∠A + ∠B)

⇒ 3y = 2(x + y)

⇒ 3y = 2x + 2y

⇒ 2x - y = 0 .... (i)

We know that the sum of the angles of a triangle is 180°.

Therefore,

∠A + ∠B + ∠C = 180°

⇒ x + y + 3y = 180

⇒ x + 4y = 180°  .....(ii)

On multiplying (i) by 4 and adding the result with (ii), we get

⇒ 8x + x = 180

⇒ 9x = 180

⇒ x = 180/9

⇒ x = 20

Putting x's value in Eq (i), we get

⇒ 2x - y = 0

⇒ 2(20) - y = 0

⇒ 40 - y = 0

⇒ y = 40

Hence, ∠A = x = 20°

∠B = y = 40°

∠C = 3y = 3(40) = 80°