c Consider the supplementary angles to explain your operations in Part B.
D
d Considering how you found the angle measures in Part B, substitute x for m∠ 1 and m∠ 2.
A
a
Triangle I
Triangle II
Triangle III
B
bTable for ∠ 1
m∠ 1
m∠ 5
m∠ 4
m∠ 3
Triangle I
140^(∘)
40^(∘)
40^(∘)
100^(∘)
Triangle II
130^(∘)
50^(∘)
50^(∘)
80^(∘)
Triangle III
120^(∘)
60^(∘)
60^(∘)
60^(∘)
Table for ∠ 2
m∠ 2
m∠ 3
m∠ 4
m∠ 5
Triangle I
80^(∘)
100^(∘)
40^(∘)
40^(∘)
Triangle II
100^(∘)
80^(∘)
50^(∘)
50^(∘)
Triangle III
120^(∘)
60^(∘)
60^(∘)
60^(∘)
C
c See solution.
D
dTable for ∠ 1
m∠ 1
m∠ 5
m∠ 4
m∠ 3
x
180^(∘)-x
180^(∘)-x
2x-180^(∘)
Table for ∠ 2
m∠ 2
m∠ 3
m∠ 4
m∠ 5
x
180^(∘)-x
x/2
x/2
Practice makes perfect
a We will first plot two points A and B on the coordinate plane. Then we will connect these points by a ruler to draw a straight line extending it to the right. Let's do it!
Next, we will draw an angle of 40^(∘) from point A.
Now we will draw the same angle measure from point B.
Finally, we can mark the point where two lines meet as C and remove the unnecessary parts.
Proceeding in the same way, we can draw two more isosceles triangle with the base angles of 50^(∘) and 60^(∘).
Triangle I
Triangle II
Triangle III
b Let's measure ∠ 1 of Triangle I using a protractor.
We can see that the measure of ∠ 1 is 140^(∘). With this, we will find the angle measures of the other angles.
m∠ 1
m∠ 5
m∠ 4
m∠ 3
140^(∘)
m∠ 5=180^(∘)-m∠ 1
m∠ 5=40^(∘)
m∠ 4=m∠ 5
m∠ 4=40^(∘)
m∠ 3=180^(∘)-(m∠ 4+m∠ 5)
m∠ 3=100^(∘)
Next, we will measure ∠ 2.
We can see that the measure of ∠ 2 is 80^(∘). Let's use the measures of the other angles using m∠2.
m∠ 2
m∠ 3
m∠ 4
m∠ 5
80^(∘)
m∠ 3=180^(∘)-m∠ 2
m∠ 3=100^(∘)
m∠ 4=m∠2/2
m∠ 4=40^(∘)
m∠ 5=m∠4
m∠ 5=40^(∘)
Using the same way, we can calculate the angles measure of the other triangles. Let's start with using ∠ 1.
m∠ 1
m∠ 5
m∠ 4
m∠ 3
Triangle I
140^(∘)
m∠ 5=180^(∘)-m∠ 1
m∠ 5=40^(∘)
m∠ 4=m∠ 5
m∠ 4=40^(∘)
m∠ 3=180^(∘)-(m∠ 4+m∠ 5)
m∠ 3=100^(∘)
Triangle II
130^(∘)
m∠ 5=180^(∘)-m∠ 1
m∠ 5=50^(∘)
m∠ 4=m∠ 5
m∠ 4=50^(∘)
m∠ 3=180^(∘)-(m∠ 4+m∠ 5)
m∠ 3=80^(∘)
Triangle III
120^(∘)
m∠ 5=180^(∘)-m∠ 1
m∠ 5=60^(∘)
m∠ 4=m∠ 5
m∠ 4=60^(∘)
m∠ 3=180^(∘)-(m∠ 4+m∠ 5)
m∠ 3=60^(∘)
Now we will do the same thing for ∠ 2.
m∠ 2
m∠ 3
m∠ 4
m∠ 5
Triangle I
80^(∘)
m∠ 3=180^(∘)-m∠ 2
m∠ 3=100^(∘)
m∠ 4=m∠2/2
m∠ 4=40^(∘)
m∠ 5=m∠4
m∠ 5=40^(∘)
Triangle II
100^(∘)
m∠ 3=180^(∘)-m∠ 2
m∠ 3=80^(∘)
m∠ 4=m∠2/2
m∠ 4=50^(∘)
m∠ 5=m∠4
m∠ 5=50^(∘)
Triangle III
120^(∘)
m∠ 3=180^(∘)-m∠ 2
m∠ 3=60^(∘)
m∠ 4=m∠2/2
m∠ 4=60^(∘)
m∠ 5=m∠4
m∠ 5=60^(∘)
c ∠ 5 is supplementary to ∠ 1, so m∠ 5=180^(∘)-m∠1. The sum of the angle measures in a triangle must be 180^(∘), so m∠ 3=180^(∘)-(m∠4+m∠5). ∠ 2 is supplementary to ∠ 3, so m∠ 3=180^(∘)-m∠2. m∠ 2 is twice as much as m∠ 4 and m∠ 5. Thus, m∠4=m∠5= m∠22.
d Let's substitute x for m∠1 into the table in part B.
