Question

Unit 8 right triangles and trigonometry homework 2 special right triangles answer key.

Answer 1

The special triangles have sin(60°) = cos(30°) =  $\frac{\sqrt{3} }{2}$, and sin(30°) =

cos(60°) = 0.5, from which we have;

1. x = 13, y = 13·√2
2. x = 15·√2, y = 15·√2
3. x = 6, y = 3·√3
4. x = 17·√(3), y = 17
5. x = y = 10
6. x = 50, y = 25
7. x = 2·√7, y = 2·√7
8. x = 16·√3, y = 8·√3
9. x = 11·√3, y = 33
10. x = 3·√2, y = 2·√6
11. x = √(10), y = 2·√(10)
12. y = 8·√7, x = 4·√7
13. x = 17·√3, y = 34·√2, z = 34,
14. x = 18·√3, y = 18, z = 9
15. x = 14·√2, y = 14, z = 14·√3
16. x = 8·√3, y = z, = 12·√2
17. x = 26·√3, y = 13·√3, z = 39·√2
18. x = $6\frac{2}{3} \cdot \sqrt{3}$, y = $3\frac{1}{3} \cdot \sqrt{3}$,  z = 10·√2
19. x = 6, y = 12, z = 12·√2
20. x = 20·√3, y = 30, z = 10·√3
21. Perimeter = 24·√5
22. Perimeter = 56·√2
23. Length of he ramp = 75 inches
24. The speed of the ball 75·√2 feet/s

Which method is used to solve special triangles?

The measures of the sides are;

1. x = 13, y = 13·√2

2. x = y, and x·√2 = 30

Which gives;

x = 15·√2, y = 15·√2

3. x = 3 ÷ 0.5 = 6, y = 3·√3

4. y = 34 × 0.5 = 17, x = 17·√(3)

5. x = y = 10

6. x = 50, y = 25

7. x·√2 = 2·√(14), which gives;

x = 2·√(14) ÷ √2 = √(28) = 2·√7 = y

x = 2·√7, y = 2·√7

8. x = 24 × 2 ÷ √3 = 16·√3

y = 8·√3

9. x = 11·√3, y = 33

10. y = 2·√6, x = 3·√2

11. x = √(10), y = 2·√(10)

12. y = 4·√(21) × 2 ÷ √3 = 8·√7

x = 4·√7

13. x = 17 ÷ tan(30°) = 17·√3

Common side = 34 = z

y = 34·√2

14. x = 27 × 2 ÷ √3 = 54·√3 ÷ 3 = 18·√3

Common side = 9·√3

y = 9·√3 × 2 ÷ √3 = 18

z = 9

15. x = 14·√2

Common side = 28

y = 14, z = 14·√3

16. x = 8·√3

Common side = 24

y·√2 = 24

Therefore;

y = 12·√2 = z

17. Common side = 39

x = 39 × 2 ÷ √3 = 26·√3

y = 13·√3

z = 39·√2

18. z = 10·√2 = The common side

x = 10·√2 × 2 ÷ √3 =  20·√6 ÷ 3

x = $\underline{6\frac{2}{3} \cdot \sqrt{3}}$

y = $\underline{3\frac{1}{3} \cdot \sqrt{3}}$

19. The common side = 12 = y

x = 6, z = 12·√2

20. z = 10·√3, x = 20·√3, y = 30

21. The perimeter = 3 × 8·√(5) = 24·√5

22. The perimeter = 4 × 14·√2 = 56·√2

23. The length of the ramp = 2 × 37.5 inches = 75 inches

24. Distance from the first base to the third base = 90·√2 feet

$Speed = \dfrac{90 \cdot \sqrt{2} }{1.2} = 75 \cdot \sqrt{2}$

The speed of the ball 75·√2 feet/s

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