Download Section 2.6 Problem Solving in Geometry

Math 123 - Section 2.6 - Problem Solving in Geometry - Part II - Page 1
Section 2.6
Problem Solving in Geometry - Part II
I.
Angle Problems
A.
The sum of the measures of the interior angles of a triangle is 180o.
B.
Examples – Find the measure of each angle in the triangles described.
1.
One angle of a triangle is three times as large as another. The measure of
the third angle is 30o greater than that of the smallest angle. Find the
measure of each angle. (Page 178, #36)
The first thing that we need to know is that the sum of all the interior angles of
a triangle is 180o. Now we can define our variables.
Let x = The measure of the smallest angle.
Let 3x = The measure of the second angle.
Let x + 30 = The measure of the third angle.
1st angle + 2nd angle + 3rd angle = 180
x + 3x + x + 30 = 180
5x + 30 = 180
5x = 150
x = 30
3x = 3(30) = 90
x + 30 = 30 + 30 = 60
Answer:
2.
The angles measure 30o, 60o, and 90o.
Equation: x + 2x + x + 20 = 180; The angles are 40o, 80o, and 60o.
Complementary and Supplementary Angles
1.
Two angles are complementary angles if the sum of their measures is 90o.
2.
Two angles are supplementary angles if the sum of their measures is 180o.
3.
Let’s look at some specific examples and see if we can detect a pattern here.
Angle
48o
52o
89o
xo
C.
Answer the question.
Now you try one: One angle of a triangle is twice as large as another. the
measure of the third angle is 20o more than that of the smallest angle. Find
the measure of each angle. (Page 178, #35)
Answer:
B.
Combine like terms to simplify the left-hand side.
Subtract 30 from both sides to isolate the variable term.
Divide both sides by 5 to isolate the variable.
Substitute to find the other angles.
Complement
90o – 48o = 42o
90o – 52o = 38o
90o – 89o = 1o
90o – xo
Supplement
180o – 48o = 132o
180o – 52o = 128o
180o – 89o = 91o
180o – xo
Examples – Use the five-step problem-solving strategy to find the measure of the
angle described.
1.
The angle’s measure is 78o less than that of its complement. (Page 178, #46)
First, we note that “less than” means subtract, switching the order!
© Copyright 2012 by John Fetcho. All rights reserved
Math 123 - Section 2.6 - Problem Solving in Geometry - Part II - Page 2
Next, since complement is what we are working with, the sum of the two
angles is 90o. Finally, remember that “is” means =. Now we need to define
our variables.
Let x = The measure of the angle.
Let 90 – x = The measure of its complement.
x = (90 – x) – 78 Combine like terms on the right-hand side to simplify.
x = 12 – x
Add x to both sides to get the variable terms on the same side.
2x = 12
Divide both sides by 2 to isolate the variable.
x=6
Answer the question.
Answer:
The angle measures 6o.
2.
Now you try one: The measure of the angle’s supplement is 10o more than
three times that of its complement. (Page 178, #49)
Answer:
Equation: 180 – x = 10 + 3(90 – x); The angle measures 50o.
© Copyright 2012 by John Fetcho. All rights reserved