Hi!
Let's start with some facts to help us solve:
All triangles have interior angles that add up to 180 degrees.Supplementary angles also add up to 180 degrees - they are angles on a straight line.
PROBLEM NUMBER ONE:
We can solve for angle HKG by using supplementary angles. The angle next to it and itself will add up to 180 degrees, as they are on a straight line.
[tex]120 + ?=180[/tex]
[tex]?=60[/tex]
Angle HKG is 60 degrees.
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Now, we also know that in triangles, the angles are 180 degrees. We can add up all the angles and find x:
[tex]60+(6x-10)+(2x+10)=180[/tex]
Remove parenthesis:
[tex]60+6x-10+2x+10=180[/tex]
Add like terms:
[tex]8x+60=180[/tex]
Subtract 60 from both sides:
[tex]8x=120[/tex]
Divide both sides by 8:
[tex]x=15[/tex]
PROBLEM NUMBER TWO
We have more supplementary angles. We can find angle JHG:
[tex]?+72=180[/tex]
[tex]?=108[/tex]
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Now add up the interior angles of the triangle:
[tex]4x-2+3x+4+108=180[/tex]
Add like terms:
[tex]7x+110=180[/tex]
Subtract 110 from both sides:
[tex]7x=70[/tex]
Divide both sides by 7:
[tex]x=10[/tex]
PROBLEM NUMBER TWO
We have more supplementary angles. We can find angle BDC:
[tex]?+111=180[/tex]
[tex]?=69[/tex]
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Now add up the interior angles of the triangle:
[tex]69+7x+5+8x-14=180[/tex]
Add like terms:
[tex]60+15x=180[/tex]
Subtract 60 from both sides:
[tex]15x=120[/tex]
Divide both sides by 15:
[tex]x=8[/tex]
