Me find the rest of the angles. the big triangle problem

∠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:

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°

use the power rule  

( a b ) n = a n b n

to distribute the exponent.

apply the product rule to  

3 x 2 y 3 . ( 3 x 2 ) 2 ( y 3 ) 2

apply the product rule to  

3 x 2 . 3 2 ( x 2 ) 2 ( y 3 ) 2

raise  

3

to the power of  

2 . 9 ( x 2 ) 2 ( y 3 ) 2 multiply the exponents in   ( x 2 ) 2

.

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 2 ⋅ 2 ( y 3 ) 2

multiply  

2   by   2 . 9 x 4 ( y 3 ) 2

multiply the exponents in  

( y 3 ) 2 .

apply the power rule and multiply exponents,  

( a m ) n = a m n . 9 x 4 y 3

⋅

2

multiply  

3   by   2 . 9 x 4 y 6

step-by-step explanation:

g = (2, 2)

step-by-step explanation:

given the ratio is 1 : 1 then f is the midpoint of eg

let the coordinates of g = (x, y)

using the midpoint formula

(0 + x) = 1 ( multiply both sides by 2 )

0 + x = 2 ⇒ x = 2

similarly

(4 + y) = 3 ( multiply both sides by 2 )

4 + y = 6 ( subtract 4 from both sides )

y = 2

coordinates of g = (2, 2)