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 or explain why not?

1. The artist can form just one unique triangle using the given measurements. To solve the triangle, we can use the Law of Sines or the Law of Cosines.

Using the Law of Sines:
sin(A) / a = sin(B) / b
sin(39°) / 4.2 = sin(B) / 5.7
sin(B) = (5.7 * sin(39°)) / 4.2
B ≈ sin^(-1)((5.7 * sin(39°)) / 4.2)

Using the Law of Cosines:
c^2 = a^2 + b^2 - 2ab * cos(C)
c^2 = 4.2^2 + 5.7^2 - 2 * 4.2 * 5.7 * cos(39°)
c ≈ sqrt(4.2^2 + 5.7^2 - 2 * 4.2 * 5.7 * cos(39°))

2. To find the sine and cosine of angles, you can reference the special angles in the first quadrant which lie on the unit circle.
a. For 330 degrees, you can reference the angle of 30 degrees.
b. For 120 degrees, you can reference the angle of 60 degrees.
c. For 113 degrees, there is no special angle in the first quadrant to reference.
d. For 203 degrees, you can reference the angle of 23 degrees.

3. To solve for x in the given equation:
(3/(x+1)) + (4/x) = 2
First, let's find a common denominator:
(3 * x) / (x * (x+1)) + (4 * (x+1)) / (x * (x+1)) = 2
Simplifying the equation:
(3x + 4(x+1)) / (x * (x+1)) = 2
(3x + 4x + 4) / (x * (x+1)) = 2
(7x + 4) / (x * (x+1)) = 2
Cross-multiplying:
7x + 4 = 2x * (x+1)
7x + 4 = 2x^2 + 2x
2x^2 + 2x - 7x - 4 = 0
2x^2 - 5x - 4 = 0
Now you can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Once you find the values for x, you can round them to the nearest tenth if required.

4. Sketching a unit circle with different angles:
a. An angle in the second quadrant, such as 135 degrees, will have a positive cosine and a negative sine.
b. An angle in the third quadrant, such as 270 degrees, will have a sine of -1.
c. An angle in the third quadrant, such as 230 degrees, will have a negative cosine and a negative sine.
d. An angle with a cosine of about -0.9 and a sine of about 0.4 could be around 155 degrees.
e. No angle can have a sine equal to 0.9 and cosine equal to 0.8 on the unit circle, as the sum of their squares cannot exceed 1.

1. To determine the number of different triangles that can be formed using the given measurements, we need to consider the triangle inequality theorem. According to this theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

Let's label the given measurements as follows:
a = 4.2 cm
b = 5.7 cm
angle A = 39 degrees

Using the triangle inequality theorem, we can determine the possible range of values for the third side, c:
b - a < c < b + a

Substituting the given values, we have:
5.7 cm - 4.2 cm < c < 5.7 cm + 4.2 cm
1.5 cm < c < 9.9 cm

Therefore, the possible range of values for side c is 1.5 cm to 9.9 cm.

Now, let's solve the triangle. We can use the Law of Cosines to find the missing angle and the lengths of the sides.

c^2 = a^2 + b^2 - 2ab*cos(A)

Substituting the given values, we have:
c^2 = (4.2 cm)^2 + (5.7 cm)^2 - 2(4.2 cm)(5.7 cm)*cos(39 degrees)

Calculating the right side of the equation, we get:
c^2 ≈ 91.5 cm^2

Taking the square root of both sides, we find:
c ≈ 9.6 cm (rounded to the nearest tenth)

So, there is one possible triangle that can be formed using the given measurements, and the approximate length of side c is 9.6 cm.

2. In the first quadrant, all angles have positive values for both sine and cosine. Therefore, all angles can be referenced.

a. 330 degrees is not in the first quadrant. However, its reference angle in the first quadrant would be 360 degrees - 330 degrees = 30 degrees.

b. 120 degrees is in the first quadrant itself. We can directly use this angle to find the sine and cosine.

c. 113 degrees is in the first quadrant itself. We can directly use this angle to find the sine and cosine.

d. 203 degrees is not in the first quadrant. However, its reference angle in the first quadrant would be 203 degrees - 180 degrees = 23 degrees.

3. Let's solve for x in the given equation:

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

To simplify the equation, we first need to find the common denominator, which is x:
((3 + x)/x) + (4/x) = 2

Combining the fractions:
(3 + x + 4)/x = 2

Simplifying the numerator:
(x + 7)/x = 2

Cross-multiplying:
x + 7 = 2x

Subtracting x from both sides:
7 = x

Therefore, the solution is x = 7.

4. Sketch a unit circle with angles marked in degrees (0°, 30°, 45°, 60°, 90°, 120°, 135°, 150°, 180°, 210°, 225°, 240°, 270°, 300°, 315°, 330°, 360°).

a. To have a positive cosine and negative sine, we can draw an angle in the third quadrant, such as 225°.

b. To have a sine of -1, we need to draw an angle in the third quadrant, such as 270°.

c. To have a negative cosine and negative sine, we can draw an angle in the second quadrant, such as 150°.

d. To have a cosine of about -0.9 and a sine of about 0.4, we can draw an angle in the second quadrant, such as 120°.

e. An angle cannot have both a sine equal to 0.9 and a cosine equal to 0.8 on a unit circle, as the sine and cosine values are limited to the range of -1 to 1.