Question

plss anyone tell me the answer of this question​

Answer 1

Answer:

Yes I will answer

Step-by-step explanation:

wkt sum of all angles in a triangle is 180 degrees.

Therefore

x + 40 + 2x + 30 + 3x-10 =180

= 6x + 60=180

6x = 180-60=120

x = 120/6=20

Therefore, 1st angle = x+40=20+40=60

2nd angle =2x+30=2*20+30=40+30=70

3rd angle =3*20-10=60-10=50

Verification = we know that angles of three sides of triangle is 180 degrees.

So, 60+70+50=180 degrees.

Hence Proved

Answer 2

Given :

• A triangle ABC in which ∠A = x+40° , ∠B = 2x+30° and ∠C = 3x-10° .

To Find :

• All three angles of Triangle .

Solution :

As we know that Sum of all angles of a triangle is 180° . So ,

$\longmapsto\tt{\angle{A}+\angle{B}+\angle{C}=180^{\circ}}$

$\longmapsto\tt{x+40^{\circ}+2x+30^{\circ}+3x-10^{\circ}=180^{\circ}}$

$\longmapsto\tt{6x+60^{\circ}=180^{\circ}}$

$\longmapsto\tt{6x=180^{\circ}-60^{\circ}}$

$\longmapsto\tt{6x=120^{\circ}}$

$\longmapsto\tt{x=\cancel\dfrac{120}{6}}$

$\longmapsto\tt\bf{x=20}$

Value of x is 20 ..

Therefore :

$\longmapsto\tt{Measure\:of\:\angle{A}=20^{\circ}+40^{\circ}}$

$\longmapsto\tt\bf{60^{\circ}}$

$\longmapsto\tt{Measure\:of\:\angle{B}=2(20)+30^{\circ}}$

$\longmapsto\tt\bf{70^{\circ}}$

$\longmapsto\tt{Measure\:of\:\angle{C}=3(20)-10^{\circ}}$

$\longmapsto\tt\bf{50^{\circ}}$