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2 years ago

A dartboard is divided into 20 sectors. Each sector is worth a point value from 1 to 20 and has shaded regions that double or

Mathematics

1 answer:

Nikitich [7]2 years ago

8 0

The area of each sector is the amount of space on the sector

The area of the entire region is 6.89
The area of the double region is 0.75
The area of the triple region is 0.46

<h3>The area of the entire sector</h3>

The area of a sector is calculated as:

A = (1/2) × r^2θ

For the entire sector, we have:

θ = π/10

r = 6.625

So, we have:

A = (1/2) × 6.625^2 × π/10

Evaluate the product

A = 6.89

<h3>The area of the double region </h3>

For the double region, we have:

θ = π/10

r = 6.25

So, we have:

A = Complete sector - (1/2) × r^2θ

A = 6.89 - (1/2) × 6.25^2 × π/10

Evaluate the difference

A = 0.75

<h3>The area of the triple region </h3>

For the triple region, we have:

θ = π/10

R = 4.125

r = 3.75

So, we have:

A = (1/2) × R^2θ - (1/2) × r^2θ

A = (1/2) × 4.125^2 × π/10 - (1/2) × 3.75^2 × π/10

Evaluate the difference

A = 0.46

Hence, the area of the triple region is 0.46

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Use the given information to find the exact value of the trigonometric function

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$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

Also,