1. An artist is designing triangular mirrors. Determine the number of different triangles that she can form using the given measurements. Then solve the triangles. Round to the nearest tenth.

a=4.2 cm b= 5.7 cm measure angle A= 39 degrees

2. What angle in the first quadrant could you reference to help you find the sine and cosine of each of the following angles? a. 330 degrees b. 120 degrees c. 113 degrees d. 203 degrees

3. Solve for x. (3/x+1)+4/x=2

**I need help especially on this.4. Sketch a unit circle. In your circle, sketch in an angle that has:

a. A positive cosine and a negative sine. b. A sine of -1. c. A negative cosine and a negative sine. d. A cosine of about -0.9 and a sine of about 0.4. e. Could an angle have a sine equal to 0.9 and cosine equal to 0.8? Give an example angle explain why not?

1. To determine the number of different triangles that can be formed, we can use the formula: n = (n-2)(n-1)/2, where n is the number of given measurements. In this case, n = 3 (a, b, and angle A). So we have: n = (3-2)(3-1)/2 = 1. This means there is only one triangle that can be formed.

To solve the triangle:
Given measurements:
a = 4.2 cm
b = 5.7 cm
angle A = 39 degrees

Step 1: Use the Law of Sines to find angle B: sin(B)/b = sin(A)/a
sin(B)/5.7 = sin(39)/4.2
sin(B) = (5.7 * sin(39))/4.2
B ≈ arcsin((5.7 * sin(39))/4.2)

Step 2: Use the triangle angle sum property to find angle C: angle C = 180 - angle A - angle B

Step 3: Use the Law of Sines to find side c: c/sin(C) = a/sin(A)
c/sin(C) = 4.2/sin(39)
c = (4.2 * sin(C))/sin(39)

Step 4: Use the Law of Cosines to find side b: b^2 = a^2 + c^2 - 2ac * cos(B)

You can follow these steps, substitute the values, and use a calculator to solve the trigonometric functions. Round the answers to the nearest tenth.

2. To find the reference angle in the first quadrant for each given angle, subtract the angles from 180 degrees.

a. 330 degrees:
Reference angle = 180 - 330 = 150 degrees

b. 120 degrees:
Reference angle = 180 - 120 = 60 degrees

c. 113 degrees:
Reference angle = 180 - 113 = 67 degrees

d. 203 degrees:
Reference angle = 180 - 203 = -23 degrees

3. To solve for x in the given equation: (3/x+1) + (4/x) = 2

Step 1: Simplify the equation:
[(3 * (x+1)) + (4 * x)] / x(x+1) = 2
(3x + 3 + 4x) / x(x+1) = 2
(7x + 3) / x(x+1) = 2

Step 2: Multiply both sides of the equation by x(x+1) to eliminate the denominators:
7x + 3 = 2x^2 + 2x

Step 3: Rearrange the equation to form a quadratic equation:
2x^2 + 2x - 7x - 3 = 0
2x^2 - 5x - 3 = 0

Step 4: Solve the quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, using the quadratic formula would be simplest:
x = (-b ± √(b^2 - 4ac)) / (2a)
x = (-(-5) ± √((-5)^2 - 4(2)(-3))) / (2(2))
x = (5 ± √(25 + 24)) / 4
x = (5 ± √(49)) / 4
x = (5 ± 7) / 4

Step 5: Solve for x:
x1 = (5 + 7) / 4 = 3
x2 = (5 - 7) / 4 = -1/2

So the solutions for x are x = 3 and x = -1/2.

**Please note that it's important to check if the solutions satisfy the original equation.

1. To determine the number of different triangles that can be formed using the given measurements, we need to consider the conditions necessary for a valid triangle. In a triangle, the sum of the lengths of any two sides must be greater than the length of the third side. Let's call the given measurements a, b, and the angle A.

Using this information, we can calculate the number of triangles by checking the conditions:

Condition 1: a + b > c
Condition 2: b + c > a
Condition 3: c + a > b

Given: a = 4.2 cm, b = 5.7 cm, and Angle A = 39 degrees

We can solve for the third side, c, using the Law of Cosines:

c^2 = a^2 + b^2 - 2ab*cos(A)
c^2 = 4.2^2 + 5.7^2 - 2*4.2*5.7*cos(39)
c ≈ 4.8907 cm (rounded to the nearest tenth)

Now, let's check the conditions for a triangle:

Condition 1: 4.2 + 5.7 > 4.8907 (True)
Condition 2: 5.7 + 4.8907 > 4.2 (True)
Condition 3: 4.8907 + 4.2 > 5.7 (True)

All three conditions are satisfied, so a valid triangle can be formed.

To solve the triangle, we can use the Law of Sines and Law of Cosines. Since we have two sides and an included angle, we can use the Law of Sines to find the remaining angles:

sin(A)/a = sin(B)/b = sin(C)/c

sin(B) = (b*sin(A))/a
sin(B) = (5.7*sin(39))/4.2
B ≈ 56.7 degrees (rounded to the nearest tenth)

The sum of angles in a triangle is 180 degrees, so:

C ≈ 84.3 degrees (180 - 39 - 56.7)

We have found the measures of angles B and C. To round them to the nearest tenth, B ≈ 56.7 degrees and C ≈ 84.3 degrees.

2. To find the angle in the first quadrant that can help us find the sine and cosine of each given angle, we need to reference the reference angle. The reference angle is the acute angle between the terminal side of an angle and the x-axis.

a. For the angle 330 degrees:
The reference angle can be found by subtracting 360 degrees from 330 degrees:
Reference angle = 330 degrees - 360 degrees = -30 degrees
In the first quadrant, the angle with the same reference angle is 30 degrees. We can use the sine and cosine of 30 degrees to find the sine and cosine of 330 degrees.

b. For the angle 120 degrees:
The reference angle is found by subtracting 90 degrees from 120 degrees:
Reference angle = 120 degrees - 90 degrees = 30 degrees
The angle in the first quadrant with the same reference angle is also 30 degrees. So, we can use the sine and cosine of 30 degrees.

c. For the angle 113 degrees:
The reference angle is found by subtracting 90 degrees from 113 degrees:
Reference angle = 113 degrees - 90 degrees = 23 degrees
The angle in the first quadrant with the same reference angle is 23 degrees. We can use the sine and cosine of 23 degrees.

d. For the angle 203 degrees:
The reference angle can be found by subtracting 180 degrees from 203 degrees:
Reference angle = 203 degrees - 180 degrees = 23 degrees
Again, the angle in the first quadrant with the same reference angle is 23 degrees. So, we can use the sine and cosine of 23 degrees.

3. Let's solve the equation for x: (3/x + 1) + (4/x) = 2

To solve this equation, we need to eliminate the denominators. To do this, we can multiply every term in the equation by the least common multiple (LCM) of the denominators. In this case, the LCM is x.

x(3/x + 1) + x(4/x) = x(2)

Simplifying this expression gives us:

3 + x(4/x) + 4 = 2x

Simplifying further:

3 + 4 + 4/x = 2x

Combining like terms:

7 + 4/x = 2x

Now, we can multiply the entire equation by x to eliminate the fraction:

7x + 4 = 2x^2

Rearranging the equation, we get:

2x^2 - 7x - 4 = 0

This is now a quadratic equation. Solving for x, we can factor or use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

Using the quadratic formula with a = 2, b = -7, and c = -4, we find:

x = (7 ± √(7^2 - 4*2*(-4))) / (2*2)
x = (7 ± √(49 + 32)) / 4
x = (7 ± √81) / 4
x = (7 ± 9) / 4

x = (7 + 9) / 4 = 4
x = (7 - 9) / 4 = -1/2

So, the values of x that satisfy the equation are x = 4 and x = -1/2.

4. To sketch a unit circle:
- Draw a circle with a radius of 1 unit centered at the origin (0, 0).
- Mark the positive x-axis to the right as (1, 0).
- Mark the positive y-axis upward as (0, 1) and the negative y-axis downward as (0, -1).

a. A positive cosine and a negative sine:
- Sketch an angle in the second quadrant (between the positive y-axis and the negative x-axis) but below the x-axis.
- The cosine value will be positive because it is the x-coordinate of the point on the unit circle in this quadrant.
- The sine value will be negative because it is the y-coordinate of the point on the unit circle in this quadrant.

b. A sine of -1:
- Sketch an angle in the fourth quadrant (below the x-axis and to the right of the negative y-axis).
- The angle will have a y-coordinate of -1 since sine represents the y-coordinate of a point on the unit circle.

c. A negative cosine and a negative sine:
- Sketch an angle in the third quadrant (between the negative x-axis and the negative y-axis).
- Both the cosine and sine values will be negative because both the x-coordinate and y-coordinate will be negative.

d. A cosine of about -0.9 and a sine of about 0.4:
- To find such an angle, we need to identify a point on the unit circle with these approximate coordinates.
- The cosine value of -0.9 corresponds to the x-coordinate, and the sine value of 0.4 corresponds to the y-coordinate.
- Sketch an angle in the second quadrant that produces approximately these coordinates.

e. Could an angle have a sine equal to 0.9 and cosine equal to 0.8? Give an example angle and explain why not?
- No, an angle cannot have a sine equal to 0.9 and cosine equal to 0.8 simultaneously.
- The maximum possible magnitude for both sine and cosine is 1 because they represent ratios of sides in a right triangle.
- Therefore, it is not possible for an angle to have a sine or cosine value greater than 1.