Question

In ∆ABC , angle C = 3 angle B = 2(angle A + angle B)

Answer 1

Explanation:

Explanation:

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$\bf{QUESTION - }$

In a ΔABC , ∠C = 3∠B = 2(∠ A + ∠ B ) . Find the angles .

→ let ∠a = x° and ∠b = y°

Then

∠C= 3∠B= 3(y°)

Now,

→ ∠C = 2(∠A+∠B)

=> 3y = 2(x+y)

=> 2x - y = 0............(1)

we know that the sum of angles of triangle is 180°

.°. ∠A + ∠B + ∠C = 180°

=> x + y + 3y = 180

=> x + 4y = 180...........(2)

on multiplying (1) by 4 and adding result with (2), we get

8x + x = 180

= 9x = 180

=> x = 180/9

=> x = 20

putting x = 20 in equation (1)

we get

→ y=(2×20)

→ y=40

thus,

x=20

y=40

Hence

→ ∠a = 20°

→ ∠a = 20° → ∠b=40°

$\sf{ \angle \: c = (3 \times 40) = \: 120 \degree}$

Answer 2

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