Question

See the attachment below.

Answer 1

Answers:

a) $L=208\pi$

b) $A=0.18 cm^{2}$

Step-by-step explanation:

a) The area of the sector of a circle $A_{c}$ is given by:

$A_{c}=\frac{rL}{2}=\frac{r^{2}\theta}{2}$ (1)

Where:

$r=144$ is the radius

$L$ is the length of arc  

$\theta=260\°=\frac{13}{9}\pi$ is the angle in radians (knowing $360\°=2\pi$ to make the conversion)

Isolating $L$ from (1) :

$L=r\theta$ (2)

$L=144(\frac{13}{9}\pi)$ (3)

$L=208\pi$ (4)  This is the length of arc

b) If we want to find the shaded area $A$, we have to find the area of the sector of a circle $A_{c}$ and substract the area of the triangle $A_{t}$.

We already know $A_{c}=\frac{r^{2}\theta}{2}$ (5)

Where:

$r=4cm$

$\theta=30\°=\frac{\pi}{6}$ (remembering $360\°=2\pi$)

Hence:

$A_{c}=\frac{(4cm)^{2}(\frac{\pi}{6})}{2}$ (6)

$A_{c}=\frac{4}{3}\pi cm^{2}$ (7) area of the sector of the circle

On the other hand, the area of a triangle is given by:

$A_{t}=\frac{bh}{2}$ (8)

Where:

$b$ is the base of the triangle

$h$ is the height of the triangle

If we divide this triangle in half, we will have two right triangles, each one with a height $h$  and a base $\frac{b}{2}$, and hypotenuse=4cm.

In addition, each triangle will have the following angles (in degrees): $90\°$, $15\°$ (the half of $30\°$), $75\°$ (knowing the three inner angles of a triangle sum to $180\°$).

Having this clear, let's use the trigonometric functions sine and cosine to find the values of $h$ and $b$:

$sin\alpha=\frac{Oppositeside}{Hypotenuse}$ (9)

$cos\alpha=\frac{Adjacentside}{Hypotenuse}$ (10)

$sin(15\°)=\frac{\frac{b}{2}}{4cm}$ (11)

$b=8sin(15\°)=2.070cm$ (12) This is the base of the triangle

$cos(15\°)=\frac{h}{4cm}$ (13)

$h=4cos(15\°)=3.86cm$ (14)  This is the height of the triangle

Substituting (12) and (14) in (8):

$A_{t}=\frac{(2.070cm)(3.86cm)}{2}$ (15)

$A_{t}=3.99cm^{2} \approx 4cm^{2}$ (16) This is the area of the triangle

Substracting the area of the triangle from the area of the sector of the circle:

$A=A_{c}-A_{t}$ (17)

$A=\frac{4}{3}\pi cm^{2}- 4cm^{2}$ (18)

Finally we have the area of the shaded portion:

$A=0.18 cm^{2}$