find the angle of triangle (x-20),(x-40) and (x/2-10)
Answers
Answered by
12
x-20 + x-40 + x/2-10 =180 (angle sum property)
x+x+x/2-20-40-10=180
(2x+2x+x)/2--70=180
5x/2=180+70
=250
5x=250*2
5x=500
x=100
x-20=80,x-40=60,x/2-10=50-10=40
the angles of triangle are 80,60,40
x+x+x/2-20-40-10=180
(2x+2x+x)/2--70=180
5x/2=180+70
=250
5x=250*2
5x=500
x=100
x-20=80,x-40=60,x/2-10=50-10=40
the angles of triangle are 80,60,40
Answered by
6
Given angles are (x-20),(x-40),(x/2-10)
We know that the sum of all angles in a triangle =180 degrees
(x-20)+(x-40)+(x/2-10)=180
2x-70+x/2 =180
(4x-140+x)/2=180
5x-140=360
5x=500
x=100 degrees
Angles are, (x-20)=100-20=80
(x-40)=100-40=60
(x/2-10)=100/2-10=50-10=40
Therefore the angles are 80,60,40.
Similar questions