Answer:
a.
<u>cos θ = 2/5, tan θ = ?</u>
- tan θ = sin θ / cos θ
- tan θ = √sin² θ / (2/5)
- tan θ= 5√(1 - cos²θ) / 2
- tan θ = 5√(1 - 4/25) / 2
- tan θ = 5√(21/25) / 2
- tan θ = √21 / 2
b.
<u>cosec θ = 7/3, cos θ = ?</u>
- cosec θ = 1/ sin θ
- cosec θ = 1/ √(1 - cos²θ)
- √(1 - cos²θ) = 1 / cosec θ
- 1 - cos²θ = (1 / (7/3))²
- cos²θ = 1 - 9 / 49
- cos²θ = 40/49
- cos θ = √40/49
- cos θ = 2√10/7
c.
<u>cot θ = 4/3, sec θ = ?</u>
- cot θ = cos θ / sin θ
- cos θ = cot θ * sin θ
- cos θ = 4/3 * √(1 - cos²θ)
- 9cos²θ = 16(1 - cos²θ)
- 25cos²θ = 16
- cos²θ = 16/25
- cos θ √16/25
- cos θ = 4/5
- sec θ = 1/ cos θ
- sec θ = 1/ (4/5)
- sec θ = 5/4
d.
<u>tan θ = 3, cosec θ = ?</u>
- sin²θ + cos²θ = 1
- 1 + cos²θ/sin²θ = 1/ sin²θ
- 1 + 1/tan²θ = cosec²θ
- 1 + 1/9 = cosec²θ
- cosec²θ = 10/9
- cosec θ = √(10/9)
- cosec θ = √10 / 3
Take a pic of an example and I will see if I can help you
Answer:
4 5/12
Step-by-step explanation:
4 3/4 = 19/4
19/4 = 57/12
1/3 = 4/12
57/12 - 4/12 = 53/12 = 4 5/12
Answer:
the answer is 8 units so that all i have to say.
Answer:
24 meters
Step-by-step explanation:
The area of a circle can be found using the following formula.
![a=\pi r^2](https://tex.z-dn.net/?f=a%3D%5Cpi%20r%5E2)
We know that the area is 144pi m^2.
![a= 144\pi m^2](https://tex.z-dn.net/?f=a%3D%20144%5Cpi%20m%5E2)
Substitute 144pi m^2 in for a.
![144\pi m^2= \pi r^2](https://tex.z-dn.net/?f=144%5Cpi%20%20m%5E2%3D%20%5Cpi%20r%5E2)
We want to find the diameter. First, we must find the radius. We have to get the variable, r, by itself.
Divide both sides of the equation by pi.
![144\pi m^2/ \pi = \pi r^2 / \pi](https://tex.z-dn.net/?f=144%5Cpi%20%20m%5E2%2F%20%5Cpi%20%3D%20%5Cpi%20r%5E2%20%2F%20%5Cpi)
![144\pi m^2/ \pi = r^2](https://tex.z-dn.net/?f=144%5Cpi%20%20m%5E2%2F%20%5Cpi%20%3D%20%20r%5E2)
![144 m^2= r^2](https://tex.z-dn.net/?f=144%20m%5E2%3D%20r%5E2)
The variable, r , is being squared. The inverse of a square is the square root. Take the square root of both sides of the equation.
![\sqrt{144 m^2} =\sqrt{r^2}](https://tex.z-dn.net/?f=%5Csqrt%7B144%20m%5E2%7D%20%3D%5Csqrt%7Br%5E2%7D)
![\sqrt{144 m^2} =r](https://tex.z-dn.net/?f=%5Csqrt%7B144%20m%5E2%7D%20%3Dr)
![12 m= r](https://tex.z-dn.net/?f=12%20m%3D%20r)
The radius of the circle is 12 meters, but the question asks for diameter. The diameter is twice the radius.
![d= r*2](https://tex.z-dn.net/?f=d%3D%20r%2A2)
The radius is 12 m.
![d= 12 m*2](https://tex.z-dn.net/?f=d%3D%2012%20m%2A2)
Multiply
![d= 24 m](https://tex.z-dn.net/?f=d%3D%2024%20m)
The diameter of the circle is 24 meters.