BRAINLIEST ANSWER GUYS

coldgirl [10]

<h3>Answer: XWY and STR</h3>

I tend to think of parallel lines as train tracks (the metal rail part anyway). Inside the train tracks is the interior region, while outside the train tracks is the exterior region. Alternate exterior angles are found here. Specifically they are angles that are on opposite or alternate sides of the transversal cut.

Both pairs of alternate exterior angles are shown in the diagram below. They are color coded to help show how they pair up and which are congruent.

A thing to notice: choices B, C, and D all have point W as the vertex of the angles. This means that the angles somehow touch or are adjacent in some way due to this shared vertex point. However, alternate exterior angles never touch because parallel lines never do so either. We can rule out choices B,C,D from this reasoning alone. We cannot have both alternate exterior angles on the same exterior side of the train tracks. Both sides must be accounted for.

4 0

4 years ago

I answered it but I just need someone to answer just to make sure mine are right

Artemon [7]

<h2>The missing measures of the angles are:</h2>

$m\angle 1= 60^{\circ}\\\\m\angle 2= 39^{\circ}\\\\m\angle 3= 21^{\circ}\\\\m\angle 4 = 39^{\circ}\\\\m\angle 5 = 21^{\circ}$

<u>Given</u>:

$m\angle Z = 138^{\circ}$

<em><u>Note the following:</u></em>

An equilateral triangle has all its three angles equal to each other. Each angle = $60^{\circ}$ .

This implies that, since $\triangle WXY$ is equilateral, therefore, $m\angle Y = m\angle XWY = m\angle XYW = 60^{\circ}$

Base angles of an isosceles triangle are congruent to each other.

This implies that, since $\triangle WZY$ is isosceles, therefore, $m\angle 3 = \angle 5$

Applying the above stated, let's find the measure of each angle:

Find $m\angle 1$

$m\angle 1 = 60^{\circ}$ (an angle in an equilateral triangle equals 60 degrees)

Find $m\angle 2$

$m\angle 2 = 60 - m \angle 3$

$m\angle 2 = 60 -\frac{1}{2}(180 - 138)$ $(Note: \frac{1}{2}(180 - 138) = 1 $ base $ angle $ of $ \triangle WZY)$

$m\angle 2 = 60 -21\\\\m\angle 2 = 39^{\circ}$

Find $m\angle 3$ and $m\angle 5$ (base angles of isosceles triangle WZY)

$m\angle 3 = \frac{1}{2}(180 - 138) (1 $ base $ angle $ of $ \triangle WZY)$

$m\angle 3 = \frac{1}{2}(42) \\\\m\angle3 = 21^{\circ}$

$m\angle3 = m\angle 5$ (base angles of isosceles triangle are congruent)

Therefore,

$m\angle5 = 21^{\circ}$

Find $m\angle 4$

$m\angle 4 = 60 - m\angle 5$

Substitute

$m\angle 4 = 60 - 21\\\\m\angle 4 = 39^{\circ}$

The missing measures of the angles are:

$m\angle 1= 60^{\circ}\\\\ m\angle 2= 39^{\circ}\\\\m\angle 3= 21^{\circ}\\\\m\angle 4 = 39^{\circ}\\\\m\angle 5 = 21^{\circ}$

Learn more here:

brainly.com/question/2944195

5 0

3 years ago

Vernon's work for finding the value of x is shown below. Lines A C and B D intersect at point E. Angle A E D is (16 x + 8) degre

Marta_Voda [28]

Answer:

<em>No, he should have set the sum of ∠AED and ∠DEC equal to 180°, rather then setting ∠AED and ∠DEC equal to each other</em>

Step-by-step explanation:

Find the diagram attached

If line AC and BD intersects, then m<AED + m<DEC = 180 (sum of angle on a straight line is 180 degrees)

Given

m<AED = 16x+8

m<DEC = 76 degrees

16x + 8 + 76 = 180

16x + 84 = 180

16x = 180-84

16x = 96

x = 96/16

x = 6

Hence the value of x is 6

Hence the correct option is <em>No, he should have set the sum of ∠AED and ∠DEC equal to 180°, rather then setting ∠AED and ∠DEC equal to each other</em>