Chapter 5 exam Practice

Last updated over 1 year ago

7 Nsɛmmisa

Laabri

Chapter 5 exam Practice

Last updated over 1 year ago

7 Nsɛmmisa

Asemmisa {{asɛmmisaAhyɛnsode}}

Find the equation of a circle with a center of (2,3) and a radius of 3.

What are the x-intercepts of the circle?

What are the y-intercepts of the circle?

Asemmisa {{asɛmmisaAhyɛnsode}}

On a circle of radius 12 cm, find the following

a. Length of the arc that subtends a central angle of 120\degree?

b. Area of a sector with a central angle of 120\degree?

Asemmisa {{asɛmmisaAhyɛnsode}}

If (\theta)=\frac{2\pi}{3}, what are the exact values of the following:

\sin(\theta)=

\cos(\theta)=

\tan(\theta)=

\csc(\theta)=

\sec(\theta)=

\cot(\theta)=

Asemmisa {{asɛmmisaAhyɛnsode}}

If \cos(\theta)=\frac{2}{9} and \theta is in the fourth quadrant, calculate the following.

\sin(\theta)=

\tan(\theta)=

\csc(\theta)=

\sec(\theta)=

\cot(\theta)=

Asemmisa {{asɛmmisaAhyɛnsode}}

Prove the equation below is true
\frac{\csc^{2}(\theta)-1}{\csc^{2}(\theta)-\csc(\theta)}=1+\sin(\theta)

Asemmisa {{asɛmmisaAhyɛnsode}}

Solve for x

Asemmisa {{asɛmmisaAhyɛnsode}}

A 200 foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 15° and that the angle of depression to the bottom of the tower is 2°. How far is the person from the monument?

Asemmisa {{asɛmmisaAhyɛnsode}}

Find the equation of a circle with a center of (2,3) and a radius of 3.

What are the x-intercepts of the circle?

What are the y-intercepts of the circle?

Asemmisa {{asɛmmisaAhyɛnsode}}

On a circle of radius 12 cm, find the following

a. Length of the arc that subtends a central angle of 120\degree?

b. Area of a sector with a central angle of 120\degree?

Asemmisa {{asɛmmisaAhyɛnsode}}

If (\theta)=\frac{2\pi}{3}, what are the exact values of the following:

\sin(\theta)=

\cos(\theta)=

\tan(\theta)=

\csc(\theta)=

\sec(\theta)=

\cot(\theta)=

Asemmisa {{asɛmmisaAhyɛnsode}}

If \cos(\theta)=\frac{2}{9} and \theta is in the fourth quadrant, calculate the following.

\sin(\theta)=

\tan(\theta)=

\csc(\theta)=

\sec(\theta)=

\cot(\theta)=

Asemmisa {{asɛmmisaAhyɛnsode}}

Prove the equation below is true
\frac{\csc^{2}(\theta)-1}{\csc^{2}(\theta)-\csc(\theta)}=1+\sin(\theta)

Asemmisa {{asɛmmisaAhyɛnsode}}

Solve for x

Asemmisa {{asɛmmisaAhyɛnsode}}

A 200 foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 15° and that the angle of depression to the bottom of the tower is 2°. How far is the person from the monument?

Chapter 5 exam Practice

Chapter 5 exam Practice

Prove the equation below is true\frac{\csc^{2}(\theta)-1}{\csc^{2}(\theta)-\csc(\theta)}=1+\sin(\theta)

Solve for x

A 200 foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 15° and that the angle of depression to the bottom of the tower is 2°. How far is the person from the monument?

Prove the equation below is true\frac{\csc^{2}(\theta)-1}{\csc^{2}(\theta)-\csc(\theta)}=1+\sin(\theta)

Solve for x

A 200 foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 15° and that the angle of depression to the bottom of the tower is 2°. How far is the person from the monument?

Prove the equation below is true
\frac{\csc^{2}(\theta)-1}{\csc^{2}(\theta)-\csc(\theta)}=1+\sin(\theta)

Prove the equation below is true
\frac{\csc^{2}(\theta)-1}{\csc^{2}(\theta)-\csc(\theta)}=1+\sin(\theta)