Core Connections Integrated II, 2015
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Core Connections Integrated II, 2015 View details
2. Section 5.2
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Exercise 104 Page 297

Practice makes perfect
a Examining the diagram, we see that it consists of two right triangles.
In the smaller triangle, both legs are known. With this information, we can calculate this triangle's hypotenuse which also happens to be a leg in the larger triangle.
a^2+b^2=c^2
3^2+ 4^2=c^2
Solve for c
9+16=c^2
25=c^2
c^2=25
c=± 5

c > 0

c=5
The hypotenuse of the smaller triangle is 5 m. Let's add this to the diagram.
Now we have enough information to calculate the hypotenuse of the larger triangle.
a^2+b^2=c^2

a= 5, b= 12, c= x

5^2+ 12^2= x^2
Solve for x
25+144=x^2
169=x^2
x^2=169
x=± 13

x > 0

x=13
The side labeled x is 13 meters long.
b Using the Triangle Angle Sum Theorem, we know that the angle measures of a triangle sum to 180^(∘). If we call the triangle's unknown angle y, we can write an equation.
35^(∘)+65^(∘)+y=180^(∘) Let's solve this equation for y.
35+65+y=180
100+y=180
y=80
Let's add the triangle's third angle to the diagram.

To find x, we will highlight a couple of sides and a segment in our diagram.

The orange side is a transversal to the blue lines. Since the 80^(∘)-angle and x are between the lines but on opposite side of the transversal, they are alternate interior angles. Additionally, because the lines are parallel, we can say by the Alternate Interior Angles Theorem that they are congruent.

Therefore, we know that x=80^(∘).