Answer :
Great! Let's address each question step-by-step.
### Question 29:
Which values of pairs of trigonometric functions is a correct match?
A. [tex]\(\cos 10^{\circ} = \sin 80^{\circ}\)[/tex]
C. [tex]\(\cot 30^{\circ} = \sec 60^{\circ}\)[/tex]
B. [tex]\(\csc 20^{\circ} = \sin 70^{\circ}\)[/tex]
D. [tex]\(\cos 40^{\circ} = \sec 50^{\circ}\)[/tex]
We'll evaluate each pair:
A. [tex]\(\cos 10^{\circ} = \sin 80^{\circ}\)[/tex]
Using the identity [tex]\(\sin(90^\circ - \theta) = \cos(\theta)\)[/tex], we know [tex]\(\sin 80^\circ = \cos 10^\circ\)[/tex]. Therefore, option A is correct.
C. [tex]\(\cot 30^{\circ} = \sec 60^{\circ}\)[/tex]
[tex]\(\cot 30^\circ = \frac{1}{\tan 30^\circ} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3}\)[/tex]
[tex]\(\sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{\frac{1}{2}} = 2\)[/tex]
Since [tex]\(\sqrt{3} \neq 2\)[/tex], option C is incorrect.
B. [tex]\(\csc 20^{\circ} = \sin 70^{\circ}\)[/tex]
[tex]\(\csc 20^\circ = \frac{1}{\sin 20^\circ}\)[/tex]
However, [tex]\(\sin 70^\circ \neq \frac{1}{\sin 20^\circ}\)[/tex], so option B is incorrect.
D. [tex]\(\cos 40^{\circ} = \sec 50^{\circ}\)[/tex]
[tex]\(\cos 40^\circ\)[/tex]
[tex]\(\sec 50^\circ = \frac{1}{\cos 50^\circ}\)[/tex]
We need to check if [tex]\(\cos 40^\circ = \frac{1}{\cos 50^\circ}\)[/tex].
Using the identity [tex]\(\cos(90^\circ - \theta) = \sin(\theta)\)[/tex], we have [tex]\(\cos 50^\circ = \sin 40^\circ\)[/tex]. Therefore, [tex]\(\cos 40^\circ = \frac{1}{\sin 40^\circ}\)[/tex], which is not true.
So, the correct answer is A.
### Question 30:
What is the area of a 12-sided regular polygon inscribed in a circle of radius 3 cm?
To find the area of a regular [tex]\(n\)[/tex]-sided polygon inscribed in a circle of radius [tex]\(r\)[/tex], we use the formula:
[tex]\[ \text{Area} = \frac{1}{2} n r^2 \sin\left(\frac{2\pi}{n}\right) \][/tex]
For [tex]\(n = 12\)[/tex] and [tex]\(r = 3\)[/tex]:
[tex]\[ \text{Area} = \frac{1}{2} \cdot 12 \cdot 3^2 \cdot \sin\left(\frac{2\pi}{12}\right) \][/tex]
[tex]\[ = 6 \cdot 9 \cdot \sin\left(\frac{\pi}{6}\right) \][/tex]
[tex]\[ = 54 \cdot \frac{1}{2} \][/tex]
[tex]\[ = 27 \, \text{cm}^2 \][/tex]
So the correct answer is C.
### Question 31:
A container shaped in the form of a right rectangular prism has a base of 3 meters by 5 meters and a height of 2 meters. What is its total volume in [tex]\(m^3\)[/tex]?
The volume [tex]\(V\)[/tex] of a rectangular prism is given by the product of its base area and height:
[tex]\[ V = \text{Length} \times \text{Width} \times \text{Height} \][/tex]
[tex]\[ V = 3 \times 5 \times 2 \][/tex]
[tex]\[ V = 30 \, \text{m}^3 \][/tex]
So the correct answer is B.
### Question 32:
If a spherical gas balloon of volume [tex]\(V\)[/tex] has its radius [tex]\(r\)[/tex] doubled, what will happen to its volume?
The volume [tex]\(V\)[/tex] of a sphere is given by:
[tex]\[ V = \frac{4}{3} \pi r^3 \][/tex]
If the radius [tex]\(r\)[/tex] is doubled, the new radius [tex]\(r_{\text{new}}\)[/tex] is:
[tex]\[ r_{\text{new}} = 2r \][/tex]
The new volume [tex]\(V_{\text{new}}\)[/tex] is:
[tex]\[ V_{\text{new}} = \frac{4}{3} \pi (2r)^3 \][/tex]
[tex]\[ = \frac{4}{3} \pi (8r^3) \][/tex]
[tex]\[ = 8 \left( \frac{4}{3} \pi r^3 \right) \][/tex]
[tex]\[ = 8V \][/tex]
So the correct answer is A.
### Summary:
29. A ([tex]\(\cos 10^{\circ}=\sin 80^{\circ}\)[/tex])
30. C ([tex]\(27 \, \text{cm}^2\)[/tex])
31. B ([tex]\(30 \, \text{m}^3\)[/tex])
32. A ([tex]\(8V\)[/tex])
### Question 29:
Which values of pairs of trigonometric functions is a correct match?
A. [tex]\(\cos 10^{\circ} = \sin 80^{\circ}\)[/tex]
C. [tex]\(\cot 30^{\circ} = \sec 60^{\circ}\)[/tex]
B. [tex]\(\csc 20^{\circ} = \sin 70^{\circ}\)[/tex]
D. [tex]\(\cos 40^{\circ} = \sec 50^{\circ}\)[/tex]
We'll evaluate each pair:
A. [tex]\(\cos 10^{\circ} = \sin 80^{\circ}\)[/tex]
Using the identity [tex]\(\sin(90^\circ - \theta) = \cos(\theta)\)[/tex], we know [tex]\(\sin 80^\circ = \cos 10^\circ\)[/tex]. Therefore, option A is correct.
C. [tex]\(\cot 30^{\circ} = \sec 60^{\circ}\)[/tex]
[tex]\(\cot 30^\circ = \frac{1}{\tan 30^\circ} = \frac{1}{\frac{1}{\sqrt{3}}} = \sqrt{3}\)[/tex]
[tex]\(\sec 60^\circ = \frac{1}{\cos 60^\circ} = \frac{1}{\frac{1}{2}} = 2\)[/tex]
Since [tex]\(\sqrt{3} \neq 2\)[/tex], option C is incorrect.
B. [tex]\(\csc 20^{\circ} = \sin 70^{\circ}\)[/tex]
[tex]\(\csc 20^\circ = \frac{1}{\sin 20^\circ}\)[/tex]
However, [tex]\(\sin 70^\circ \neq \frac{1}{\sin 20^\circ}\)[/tex], so option B is incorrect.
D. [tex]\(\cos 40^{\circ} = \sec 50^{\circ}\)[/tex]
[tex]\(\cos 40^\circ\)[/tex]
[tex]\(\sec 50^\circ = \frac{1}{\cos 50^\circ}\)[/tex]
We need to check if [tex]\(\cos 40^\circ = \frac{1}{\cos 50^\circ}\)[/tex].
Using the identity [tex]\(\cos(90^\circ - \theta) = \sin(\theta)\)[/tex], we have [tex]\(\cos 50^\circ = \sin 40^\circ\)[/tex]. Therefore, [tex]\(\cos 40^\circ = \frac{1}{\sin 40^\circ}\)[/tex], which is not true.
So, the correct answer is A.
### Question 30:
What is the area of a 12-sided regular polygon inscribed in a circle of radius 3 cm?
To find the area of a regular [tex]\(n\)[/tex]-sided polygon inscribed in a circle of radius [tex]\(r\)[/tex], we use the formula:
[tex]\[ \text{Area} = \frac{1}{2} n r^2 \sin\left(\frac{2\pi}{n}\right) \][/tex]
For [tex]\(n = 12\)[/tex] and [tex]\(r = 3\)[/tex]:
[tex]\[ \text{Area} = \frac{1}{2} \cdot 12 \cdot 3^2 \cdot \sin\left(\frac{2\pi}{12}\right) \][/tex]
[tex]\[ = 6 \cdot 9 \cdot \sin\left(\frac{\pi}{6}\right) \][/tex]
[tex]\[ = 54 \cdot \frac{1}{2} \][/tex]
[tex]\[ = 27 \, \text{cm}^2 \][/tex]
So the correct answer is C.
### Question 31:
A container shaped in the form of a right rectangular prism has a base of 3 meters by 5 meters and a height of 2 meters. What is its total volume in [tex]\(m^3\)[/tex]?
The volume [tex]\(V\)[/tex] of a rectangular prism is given by the product of its base area and height:
[tex]\[ V = \text{Length} \times \text{Width} \times \text{Height} \][/tex]
[tex]\[ V = 3 \times 5 \times 2 \][/tex]
[tex]\[ V = 30 \, \text{m}^3 \][/tex]
So the correct answer is B.
### Question 32:
If a spherical gas balloon of volume [tex]\(V\)[/tex] has its radius [tex]\(r\)[/tex] doubled, what will happen to its volume?
The volume [tex]\(V\)[/tex] of a sphere is given by:
[tex]\[ V = \frac{4}{3} \pi r^3 \][/tex]
If the radius [tex]\(r\)[/tex] is doubled, the new radius [tex]\(r_{\text{new}}\)[/tex] is:
[tex]\[ r_{\text{new}} = 2r \][/tex]
The new volume [tex]\(V_{\text{new}}\)[/tex] is:
[tex]\[ V_{\text{new}} = \frac{4}{3} \pi (2r)^3 \][/tex]
[tex]\[ = \frac{4}{3} \pi (8r^3) \][/tex]
[tex]\[ = 8 \left( \frac{4}{3} \pi r^3 \right) \][/tex]
[tex]\[ = 8V \][/tex]
So the correct answer is A.
### Summary:
29. A ([tex]\(\cos 10^{\circ}=\sin 80^{\circ}\)[/tex])
30. C ([tex]\(27 \, \text{cm}^2\)[/tex])
31. B ([tex]\(30 \, \text{m}^3\)[/tex])
32. A ([tex]\(8V\)[/tex])
