Question

Gina wilson unit 8 homework 2
I'm asking for a whole assignment but my teacher didn't even bother to explain this or get notes. She took us to finding missing angles to this.

Answer 1

Gina Wilson unit 8 homework 2 correct responses are;

1. x = 13, y = 13·√2
2. x = 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 = y = 2·√7
8. x = 16·√3, y = 8·√3
9. x = 11·√3, y = 33
10. x = 3·√2, y = 3·√6
11. x = √10, y = 2·√10
12. x = 4·√7, y = 8·√7
13. x = 17·√3, y = 34·√2, z = 34
14. x = 18·√3, y = 18, z = 9
15. x = 14·√2, y = 7, z = 7·√3
16. x = 8·√3, y = z = 12·√2
17. x = 26·√3, y = 13·√3, z = 39·√2
18. $\displaystyle x = \frac{20 \cdot \sqrt{3} }{3}$, $\displaystyle y = \frac{10 \cdot \sqrt{3} }{3}$, z = 10·√2
19. x = 6, y = 12, 12·√2
20. x = 10·√(12), y = 30, z = 5·√12
21. The perimeter of the triangle = 24·√5
22. The perimeter of the square = 56·√2
23. 6.25 feet
24. 75·√2 ft./s

Methods used to arrive at the above responses:

Background information;

The trigonometric ratios used to find the lengths of the sides of the given right triangles are;

$\displaystyle sin(30^{\circ}) = \mathbf{\frac{1}{2}}$

$\displaystyle cos(30^{\circ}) = \mathbf{\frac{\sqrt{3} }{2}}$

$\displaystyle sin(60^{\circ}) = \mathbf{\frac{\sqrt{3} }{2}}$

$\displaystyle cos(30^{\circ}) = \mathbf{\frac{\sqrt{3} }{2}}$

Solution:

1. The given triangle is a right triangle, and one of the interior angles is 45°, we have that the length of the legs are equal

Therefore, x = 13

By Pythagorean theorem, we have;

y = √(13² + 13²) = 13·√2

2. x = y

30² = x² + y² = 2·x²

x = √(30² ÷ 2) = 15·√2 = y

$3. {} \hspace {0.3 cm}\displaystyle cos(60^{\circ}) = \frac{1}{2} = \mathbf{\frac{3}{x}}$

x = 3 × 2 = 6

y = √(6² - 3²) = 3·√3

$4. \ \displaystyle sin(30^{\circ}) = \frac{1}{2} = \mathbf{\frac{y}{34}}$

34 = 2·y

y = 17

x = √(34² - 17²) = 17·√3

5. x = y

10·√2 = √(x² + y²) = √(x² + x²) = x·√2

Therefore; x = 10 = y

$6. {} \hspace{0.75cm} \displaystyle \frac{25 \cdot \sqrt{3} }{x} = \mathbf{\frac{\sqrt{3} }{2}}$

2 × 25·√3 = x·√3

x = 50

$\displaystyle \frac{y}{50} = \frac{1}{2}$

y = 25

7. x = y

x·√2 = 2·√14

x = √2·√14 = √28 =  2·√7 = y

$8. {} \hspace{0.75cm} \displaystyle \frac{24 }{x} = \frac{\sqrt{3} }{2}$

48 = x·√3

x = 16·√3

y = 8·√3

9. x = 22·√3 ÷ 2 = 11·√3

y × 2 = √3 × 22·√3 = 66

y = 66 ÷ 2 = 33

10. y = 2  × √6 = 2·√6

2·x = √3 × 2·√6

x = √3 × √6 = 3·√2

11. x = √10

y = √(10 + 10) = 2·√10

12. 4·√(21) × 2 = y·√3

y = 8·√7

x = 4·√7

13. √3 × 34 = x × 2

x = 17·√3

z = 34

y = 34·√2

14. x × √3 = 27 × 2

x = 3 × 9 × 2 ÷ √3 = 18·√3

y·√3 = 9·√3 × 2

y = 18

z = 9

15. x = 14·√2

y = 7

z × 2 = 14 ×√3

z = 7·√3

16. x = 8·√3

16·√3 × √3 = a × 2

a = 24 (The length of the side common to both triangles)

z = y

Therefore;

y·√2 = 24

y = 24 ÷ √2 = 12·√2 = z

17. z = 39·√2

x·√3 = 39 × 2

x = 26·√3

y = 13·√3

18. z·√2 = 20

z = 10·√2

x × √3 = 10 × 2

x = 20 ÷ √3

$\displaystyle x = \frac{20}{\sqrt{3} } = \underline{\frac{20 \cdot \sqrt{3} }{3}}$

y = 10 ÷ √3

$\displaystyle \underline{ y = \frac{10 \cdot \sqrt{3} }{3}}$

19. x = 6

y = 12

z = 12·√2

20. z·√2 = 10·√6

z = 5·√12

x = 10·√12

y = 10·√12 × √3 ÷ 2 = 30

21. 4·√15 × 2 = s×√3

s = 8·√5

s = The length of one side of the equilateral triangle (all angles are 60°)

The perimeter of the equilateral triangle = 3 × s = 3 × 8·√5 = 24·√5

22. s·√2 = 28

s = The length of a side of the square

s = 14·√2

The perimeter = 4 × 14·√2 = 56·√2

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

75 inches = 75 inches ÷ 12 inches/feet = 6.25 feet

24. The length of the sides of the infield of the baseball field = 90 feet

The distance from third to first base = The diagonal of the infield = 90·√2 ft.

The horizontal speed of the ball, v = 90·√2 ft. ÷ 1.2 s =  75·√2 ft./s

Learn more about trigonometric ratios here:

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