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

Answer:

3<B=2<A+2<B

<C=3<B

A+B+C=180

A+B+3B=180. (C=3B)

A+4B=180 ......................(1)

nd,

A+B+2A+2B=180

3A+3B=180......................(2)

multiply 3 by equ. (1)

3A+12B=540..................(3)

subtract 2 from 3 ,

9B=360

B=40

C=120

A=180-160=20

A=20°

B=40°

C=120°