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McGraw Hill Integrated II, 2012
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5. Isosceles and Equilateral Triangles
0. Preparing for Integrated Math II
13 Subchapters
1. Quadratic Expressions and Equations
p. 3-89
20 Subchapters
2. Quadratic Functions and Equations
p. 91-165
20 Subchapters
3. Quadratic Functions and Relations
p. 167-223
17 Subchapters
4. Exponential and Logarithmic Functions and Relations
p. 225-271
14 Subchapters
5. Reasoning and Proof
p. 273-331
12 Subchapters
6. Congruent Triangles
p. 333-401
15 Subchapters
7. Relationships in Triangles
p. 403-471
16 Subchapters
8. Quadrilaterals
p. 473-539
14 Subchapters
9. Proportions and Similarity
p. 541-615
16 Subchapters
10. Right Triangles and Trigonometry
p. 617-711
21 Subchapters
11. Circles
p. 713-801
17 Subchapters
12. Extending Surface Area and Volume
p. 803-879
19 Subchapters
13. Probability and Measurement
p. 881-941
14 Subchapters
Start
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Exercises
p. 378-382
63 Solutions
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Exercise
31
Page
380
Page
380
Hint & Answer
Solution
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Hints
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Notice that ∠ ACD and ∠ ACB form a
straight angle
.
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Check the answer
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m∠ ACB=136^(∘)
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Progress overview
In the previous exercise, we found that m∠ ACD=44^(∘).
As we can see ∠ ACD and ∠ ACB form a
straight angle
. This means that their measures sum to 180^(∘). Thus, we can find m∠ ACB using the definition of straight angle.
m∠ ACD+m∠ ACB=180
Substitute
m∠ ACD= 44
44+m∠ ACB=180
SubEqn
LHS-44=RHS-44
m∠ ACB=136
As a result, the measure of ∠ ACB is 136^(∘).
Theorems About Triangles
Level 1 exercises - Theorems About Triangles
Level 2 exercises - Theorems About Triangles
Level 3 exercises - Theorems About Triangles
Subchapter links
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Exercises
p.378-382
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