Answers for Lesson 5-5, pp. 292-294 Exercises

1.	∠3 ≅ ∠2 because they are vertical  and m∠1 > m∠3 by Corollary to the Ext. ∠ Thm. So, m∠1 > m∠2 by subst.
2.	An ext. ∠ of a  is larger than either remote int. ∠.
3.	m∠1 > m∠4 by Corollary to the Ext. ∠ Thm. and ∠4 ≅ ∠2 because if || lines, then alt. int.  are ≅.
4.	∠M, ∠L, ∠K	5. ∠D, ∠C, ∠E
6.  ∠G, ∠H, ∠I 	7. ∠A, ∠B, ∠C
8.  ∠E, ∠F, ∠D                    	9. ∠Z, ∠X, ∠Y
10.		11. 
12.		13. 
14.		15. 
16.	No; 2 + 36.
17.	Yes; 11 + 12 > 15; 12 + 15 > 11; 11 + 15 > 12.
18.	No; 8 + 10  19.
19.	Yes; 1 + 15 > 15; 15 + 15 > 1.
20.	Yes; 2 + 9 > 10; 9 + 10 > 2; 2 + 10 > 9.
21.	No;4 + 59.	22. 4 < s < 20
23.	11 < s < 21	24. 0 < s < 12
25.	5 < s < 41	26. 3 < s < 11
27.	15 < s < 55
28.	Answers may vary. Sample: If Y is the distance between 
Wichita and Topeka, then 20 < Y < 200.
29.	Let the distance between the peaks be d and the distances 
from the hiker to each of the peaks be a and b. Then
d + a > b and d + b > a. Thus, d > b — a and d > a — b.

Answers for Lesson 5-5, pp. 292-294 Exercises (cont.)

30.	a.
b.	The third side of the 1st  is longer than the third side of
the 2nd .
c.	See diagram in part (a).
d.	The included ∠ of the first  is greater than the included
∠ of the second .
31.	Answers may vary. Sample: The shortcut across the grass is shorter than the sum of the two paths.
32.	AB
33.	a. m∠OTY
b.	m∠3
c.	Base  of an isosc. ∠ are ≅.
d.	∠ Add. Post.
e.	Comparison Prop. of Ineq.
f.	Subst. (step 2) 
g.	An ext. ∠ of a  is greater than either remote int. ∠.
h. Trans. Prop. of Ineq.
34.	∠ T is the largest ∠ in PTA. Thus PA > PT because the longest side of a  is opp. the largest ∠.
35.	RS	36.   CD	37. XY	38.
39.  (2,4), (2, 5), (2,6), (3,3), (3,4), (3, 6), (3,7), (4,3), (4,4), 
(4, 5), (4, 6), (4,7), (4,8)
40. 

Answers for Lesson 5-5, pp. 292-294 Exercises (cont.)

41.
CD = AC was given so ACD is isos. by def. of isos. . 
This means mD = mCAD.Then mDAB > mCAD 
by the Comparison Prop. of Ineq. So by subst, mDAB > 
mD and by Thm. 5-11 DB > AB. Since DC + CB = DB, by subst. DC + CB > AB. Using subst. again, AC + CB > AB.