∠1 = 100°
∠2 = 56°
∠3 = 128°
∠4 = 36°
∠5 = 94°
∠6 = 40°
∠7 = 30°
∠8 = 60°
∠9 = 26°
∠10 = 38°
∠11 = 16°
∠12 = 30°
∠13 = 60°
∠14 = 110°
∠15 = 90°
∠16 = 34°
∠17 = 86°
∠18 = 52°
∠19 = 120°
∠20 = 70°
Step-by-step explanation:
Before we solve for all the angles, first we figure out what we can with the given.
Just remember the following:
The sum of all angles in any triangle is always 180°.The sum of complementary angles (two angles that form a right angle) is 90°.The sum of supplementary angles (two angles that form a straight line) is 180°An isosceles triangle which means that two sides are equal and as a result, two angles are equal.Two line segments that are perpendicular to each other create a right angle or an angle with a measure of 90°.Vertical angles are congruent.
GIVEN: ΔABC is an isosceles triangle
∠A = ∠1 = 100°
∠B = ∠6
∠B = ∠C
100° + 2(∠6) = 180°
∠6 = 80°/2
∠6 = 40°
GIVEN: ΔDEF is an equilateral triangle. We can then assume that
∠13 = 60°
∠8 = 60°
GIVEN: ∠13 is supplementary to ∠19 based on the figure
Then we can assume that:
∠13 + ∠19 = 180°
60° + ∠19 = 180°
∠19 = 120°
GIVEN: AE ⊥ EF then we can assume that the create a right angle. This means that the measure of the angle the create or m∠AEF = 90°.
Since ∠7 is complementary to the angle of the equilateral triangle ΔDEF
we can assume that
∠7 + 60° = 90 °
∠7 = 30°
GIVEN: ∠7, ∠19 and ∠12 make a triangle based on the figure:
We can then assume that:
∠7 +  ∠19 +∠12 = 180°
Based on what we solved above:
30° + 120 + ∠12 = 180°
150° + ∠12 = 180°
∠12 = 30°
GIVEN: CG ⊥ EG
We can then assume that m∠EGC = 90°
∠ECG = ∠15
∠15 = 90°
GIVEN: ∠3 = 128 and ∠4 = 36
Notice that ∠4 and the vertical angle of ∠3 make a triangle with ∠11. Because ∠3 is the vertical of the angle that completes the triangle then we know that it is equal to 128° as well.
We can then assume:
∠4 + vertical of ∠3 + ∠11 = 180°
36° + 128° + ∠11 = 180°
164° + ∠11 = 180°
∠11 = 16°
GIVEN: ∠5 = 94°
Notice that ∠5, ∠11 and ∠20 make a triangle. We already know that:
∠11 = 16° from the previous solution.
∠5 + ∠11 + ∠20 = 180°
94° + 16° + ∠20 = 180°
110° + ∠20 = 180°
∠20 = 70°
GIVEN: ∠12, ∠6 and ∠CBD , make a triangle, where:
∠6 = 60°
∠12 = 30°
∠12 + ∠6 + ∠CBD = 180°
60° + 30° + ∠CBD = 180°
90° + ∠CBD = 180°
∠CBD = 90°
Now look at the figure and you'll notice that:
∠CBD, ∠2 and ∠16 make a triangle
∠2 = 56° (given)
∠CBD = 90°
∠CBD + ∠2 + ∠16 = 180°
90 + 56 + ∠16 = 180
146 + ∠16 = 180
∠16 = 34°
GIVEN: ∠19, ∠16, and ∠CDG make a straight line
∠19 + ∠16 + ∠CDG = 180°
120° + 34° + ∠CDG = 180°
154° + ∠CDG = 180°
∠CDG = 26°
Now see that ∠CDG, ∠3 and ∠9 make a triangle:
∠CDG + ∠3 + ∠9= 180°
26 + 128 + ∠9 = 180
154 +∠9 = 180
∠9 = 26°
GIVEN: ∠20 is supplementary to ∠14 as shown on the figure.
We can assume then that:
∠20 + ∠14 = 180°
70° + ∠14 = 180°
∠14 = 110°
GIVEN: ∠3 is supplementary to ∠18, where ∠3 = 128°
∠3 + ∠18 = 180°
128 + ∠18 = 180°
∠18 = 52°
GIVEN: ∠18, ∠10 and ∠15 make a triangle
∠18 +∠10 + ∠15 = 180
52 + ∠10 + 90 = 180
142 + ∠10 = 180
∠10 = 38°
AND LASTLY: ∠4, ∠D and ∠14 make a triangle
∠4 + ∠D + ∠14 = 180°
36 + ∠D + 110 = 180
146 + ∠D = 180
∠D = 34°
Now in line GE, ∠13, ∠17 and ∠D make a straight line
∠13 + ∠17 + ∠D = 180°
60 + ∠17 + 34 = 180
94 + ∠17 = 180
∠17 = 86°
