(MATH Analysis ONLY) Chapter 5 exam

Last updated about 1 year ago

7 Nsɛmmisa

Asemmisa {{asɛmmisaAhyɛnsode}}

Laabri

(MATH Analysis ONLY) Chapter 5 exam

Last updated about 1 year ago

7 Nsɛmmisa

Asemmisa {{asɛmmisaAhyɛnsode}}

Find the equation of a circle with a center of (1,2) and a radius of \sqrt{17}.

What are the x-intercepts of the circle?

What are the y-intercepts of the circle?

Asemmisa {{asɛmmisaAhyɛnsode}}

On a circle of radius 8 cm, find the following

a. Length of the arc that subtends a central angle of 135\degree?

b. Area of a sector with a central angle of 135\degree?

Asemmisa {{asɛmmisaAhyɛnsode}}

If \theta=\frac{5\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{4}{15} 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{\tan^{2}(\theta)\cdot \cot(\theta)}{\cos(\theta)}=(1+\tan^{2}(\theta))\cdot \sin(\theta)

Asemmisa {{asɛmmisaAhyɛnsode}}

Solve for x

Asemmisa {{asɛmmisaAhyɛnsode}}

Kelly is flying a kite to which the angle of elevation is 70˚ and the hand that is holding the kite is 4 feet above the ground. The string on the kite is 65 meters long. How far is the kite above the ground?

Find the equation of a circle with a center of (1,2) and a radius of \sqrt{17}.

What are the x-intercepts of the circle?

What are the y-intercepts of the circle?

Asemmisa {{asɛmmisaAhyɛnsode}}

On a circle of radius 8 cm, find the following

a. Length of the arc that subtends a central angle of 135\degree?

b. Area of a sector with a central angle of 135\degree?

Asemmisa {{asɛmmisaAhyɛnsode}}

If \theta=\frac{5\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{4}{15} 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{\tan^{2}(\theta)\cdot \cot(\theta)}{\cos(\theta)}=(1+\tan^{2}(\theta))\cdot \sin(\theta)

Asemmisa {{asɛmmisaAhyɛnsode}}

Solve for x

Asemmisa {{asɛmmisaAhyɛnsode}}

Kelly is flying a kite to which the angle of elevation is 70˚ and the hand that is holding the kite is 4 feet above the ground. The string on the kite is 65 meters long. How far is the kite above the ground?

(MATH Analysis ONLY) Chapter 5 exam

(MATH Analysis ONLY) Chapter 5 exam

Prove the equation below is true\frac{\tan^{2}(\theta)\cdot \cot(\theta)}{\cos(\theta)}=(1+\tan^{2}(\theta))\cdot \sin(\theta)

Solve for x

Kelly is flying a kite to which the angle of elevation is 70˚ and the hand that is holding the kite is 4 feet above the ground. The string on the kite is 65 meters long. How far is the kite above the ground?

Prove the equation below is true\frac{\tan^{2}(\theta)\cdot \cot(\theta)}{\cos(\theta)}=(1+\tan^{2}(\theta))\cdot \sin(\theta)

Solve for x

Kelly is flying a kite to which the angle of elevation is 70˚ and the hand that is holding the kite is 4 feet above the ground. The string on the kite is 65 meters long. How far is the kite above the ground?

Prove the equation below is true
\frac{\tan^{2}(\theta)\cdot \cot(\theta)}{\cos(\theta)}=(1+\tan^{2}(\theta))\cdot \sin(\theta)

Prove the equation below is true
\frac{\tan^{2}(\theta)\cdot \cot(\theta)}{\cos(\theta)}=(1+\tan^{2}(\theta))\cdot \sin(\theta)