Question

The interior angles of triangles are (3x-10), (3x+10), (3x).Find the angles

Answer 1

(3x-10)+(3x+10)+3x=180° [Angle sum Property of Triangle]
=> 3x+3x+3x+10-10=180°
=>9x=180°
=>x=180/9=20°
Therefore,
the required angles are-
3x-10=3*20-10=60-10=50°
3x+10=3*20+10=60+10=70°
3x=3*20=60°

I hope that helps!!!!
Thank you.....

Answer 2

Concept we will be using:
i) Sum of the angles of a triangle is 180°.
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Solution:
The angles of the triangle are (3x-10), (3x+10) and 3x.

Therefore,
Sum of the angles must be 180°.
⇒(3x-10)+(3x+10)+3x=180
⇒3x-10+3x+10+3x=180
⇒9x=180
Divide both sides by 9:
⇒x=20


Therefore, the angles are -
3x-10= 3(20)-10=60-10=50°
3x+10=3(20)+10=60+10=70°
3x=3(20)=60°

Answer: The required angles are 50°, 70° and 60°