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
b
Table 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
d
Table 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.
