Question

98 poinis for maths solver or challenger
plz, give answer . it's urgent .


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

Best of luck .

Quality answer required .

Answer 1

$\huge \bf{ \red{answer}}$


$\bf{question - }$
In a ΔABC , ∠C = 3∠B = 2(∠ A + ∠ B ) . Find the angles .

$\bf{step \: by \: step \: explanation}$


$let \angle \: a = x \degree \: and \angle \: b \: = y \degree$


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

$\bf{ \angle \: a = 20 \degree}$
$\bf{ \angle \: b = 40 \degree}$
and

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


$\bf{thanks}$

Answer 2

Answer:

20°,40°,120°

Step-by-step explanation:

Given : ∠C = 3∠B = 2(∠A + ∠B)

⇒ 3∠B = 2∠A + 2∠B

⇒ 2∠A = 3∠B - 2∠B

⇒ 2∠A = ∠B

⇒ 2∠A - ∠B = 0    ---- (i)

We know that Sum of measures of all angles in a triangle is 180°.

⇒ ∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠B + 3∠B = 180°

⇒ ∠A + 4∠B = 180°    ---- (ii)

Multiply (i) with 4, we get

⇒ 8∠A - 4∠B = 0    ---- (iii)

On solving (i) & (iii), we get

⇒ ∠A + 4∠B = 180

⇒ 8∠A - 4∠B = 0

   -----------------------

      9∠A = 180

           ∠A = 20°

Substitute ∠A in (ii), we get

⇒ ∠A + 4∠B = 180°

⇒ 20 + 4∠B = 180°

⇒ 4∠B = 160°

⇒ ∠B = 40°

Given, ∠C = 3∠B

                 = 3(40°)

                 = 120°.

Therefore, the angles are 20°,40°,120° respectively.

Hope it helps!