Answer:
![41\text{ [units squared]}](https://tex.z-dn.net/?f=41%5Ctext%7B%20%5Bunits%20squared%5D%7D)
Step-by-step explanation:
The octagon is irregular, meaning not all sides have equal length. However, we can break it up into other shapes to find the area.
The octagon shown in the figure is a composite figure as it's composed of other shapes. In the octagon, let's break it up into:
- 4 triangles (corners)
- 3 rectangles (one in the middle, two on top after you remove triangles)
<u>Formulas</u>:
- Area of rectangle with length
and width 
: 
- Area of triangle with base
and height 
: 
<u>Area of triangles</u>:
All four triangles we broke the octagon into are congruent. Each has a base of 2 and a height of 2.
Thus, the total area of one is 
The area of all four is then 
units squared.
<u>Area of rectangles</u>:
The two smaller rectangles are also congruent. Each has a length of 3 and a width of 2. Therefore, each of them have an area of 
units squared, and the both of them have a total area of 
units squared.
The last rectangle has a width of 7 and a height of 3 for a total area of 
units squared.
Therefore, the area of the entire octagon is ![8+12+21=\boxed{41\text{ [units squared]}}](https://tex.z-dn.net/?f=8%2B12%2B21%3D%5Cboxed%7B41%5Ctext%7B%20%5Bunits%20squared%5D%7D%7D)
bearing in mind that the hypotenuse is never negative, since it's just a distance unit, so if an angle has a sine ratio of -(5/13) the negative must be the numerator, namely -5/13.
![\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}](https://tex.z-dn.net/?f=%5Cbf%20cos%5Cleft%5B%20sin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%20%5Cright%5D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Bthen%20we%20can%20say%20that%7D~%5Chfill%20%7D%7Bsin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%5Cimplies%20%5Ctheta%20%7D%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Ctextit%7Btherefore%20then%7D~%5Chfill%20%7D%7Bsin%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Bopposite%7D%7B-5%7D%7D%7B%5Cstackrel%7Bhypotenuse%7D%7B13%7D%7D%7D%5Cimpliedby%20%5Ctextit%7Blet%27s%20find%20the%20%5Cunderline%7Badjacent%7D%7D)
![\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Busing%20the%20pythagorean%20theorem%7D%20%5C%5C%5C%5C%20c%5E2%3Da%5E2%2Bb%5E2%5Cimplies%20%5Cpm%5Csqrt%7Bc%5E2-b%5E2%7D%3Da%20%5Cqquad%20%5Cbegin%7Bcases%7D%20c%3Dhypotenuse%5C%5C%20a%3Dadjacent%5C%5C%20b%3Dopposite%5C%5C%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%20%5Cpm%5Csqrt%7B13%5E2-%28-5%29%5E2%7D%3Da%5Cimplies%20%5Cpm%5Csqrt%7B144%7D%3Da%5Cimplies%20%5Cpm%2012%3Da%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20cos%5Cleft%5B%20sin%5E%7B-1%7D%5Cleft%28%20-%5Ccfrac%7B5%7D%7B13%7D%20%5Cright%29%20%5Cright%5D%5Cimplies%20cos%28%5Ctheta%20%29%3D%5Ccfrac%7B%5Cstackrel%7Badjacent%7D%7B%5Cpm%2012%7D%7D%7B13%7D)
le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.
Answer:
<em>120 degrees</em>
Step-by-step explanation:
Find the diagram attached.
From the diagram;
Interior angles are m∠BCA and m∠ABC
Exterior angle is m∠DAB
The sum of interior angle of the triangle is equal to exterior
m∠BCA +m∠ABC =m∠DAB
Given
m∠ABC = 70
m∠BCA =50
m∠DAB = 70 + 50
m∠DAB = 120 degrees
<em>Hence the measure of m∠DAB is 120 degrees</em>
Answer:
The correct answer AC = 11.52
Step-by-step explanation:
From the figure attached with this answer shows the triangle ABC.
BC=4 centimeters, m angle B=m angle c, and m angle a = 20 degrees
To find AC
From the figure we can see that D is the mid point and AD⊥BC
Then BD = CD = 4/2 = 2 cm
<BAD = 10° and <CAD = 10°
By using trigonometric ratio,
sin 10 = CD/AC = 2/AC
AC = 2/sin 10 = 2/0.1736 = 11.52
Therefore the value of AC = 11.52 cm
