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stich3 [128]
3 years ago
14

I'm not sure how to start. Can anyone help with a few letters? Once I get a start, I can finish

Mathematics
2 answers:
maks197457 [2]3 years ago
4 0
A triangle is a polygon with three edges and three vertices. It is one of the basic shapes in geometry. A triangle with vertices A, B, and C is denoted.<span>A </span>triangle<span> with all sides equal is called equilateral, a</span>triangle<span> with two sides equal is called isosceles, and a</span>triangle<span> with all sides a different length is called scalene.</span>
professor190 [17]3 years ago
4 0

Just list what we know

f=b

h=g=56

d+2f=180

d=f+c

3f+c=180

a+c+f=180

d+e=180

e=2f

a+66+f=180

a+f=114

YES

a+c+f=180

a+f=114

c=66


now rewrite everything with c in it

3f+66=180

3f=114

this also means that 2f=a (f+a=114)

f=38=b

2f=a=76=e


d+2f=180

d+76=180

d=104


now h

all angles in a quadrilateral add to 360

e+d+125+h=360

76+104+125+h=360

h=55


now I

f+h+i=180

38+55+i=180

i=87

and dunno what missing angle at right (makes up far right triangle made up of 56 and i)



list:
 

a-76

b=38

c=66

d=104

e=76

f=38

g=56

h=55


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\bf csc(\theta)=-6\implies csc(\theta)=\cfrac{\stackrel{hypotenuse}{6}}{\stackrel{opposite}{-1}}\impliedby \textit{let's find the \underline{adjacent side}}&#10;\\\\\\&#10;\textit{using the pythagorean theorem}\\\\&#10;c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a&#10;\qquad &#10;\begin{cases}&#10;c=hypotenuse\\&#10;a=adjacent\\&#10;b=opposite\\&#10;\end{cases}&#10;\\\\\\&#10;\pm\sqrt{6^2-(-1)^2}=a\implies \pm\sqrt{35}=a\implies \stackrel{IV~quadrant}{+\sqrt{35}=a}

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\bf sin(\theta)=\cfrac{opposite}{hypotenuse}&#10;\qquad\qquad &#10;cos(\theta)=\cfrac{adjacent}{hypotenuse}&#10;\\\\\\&#10;% tangent&#10;tan(\theta)=\cfrac{opposite}{adjacent}&#10;\qquad \qquad &#10;% cotangent&#10;cot(\theta)=\cfrac{adjacent}{opposite}&#10;\\\\\\&#10;% cosecant&#10;csc(\theta)=\cfrac{hypotenuse}{opposite}&#10;\qquad \qquad &#10;% secant&#10;sec(\theta)=\cfrac{hypotenuse}{adjacent}

therefore, let's just plug that on the remaining ones,

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now, let's rationalize the denominator on tangent and secant,

\bf tan(\theta)=\cfrac{-1}{\sqrt{35}}\implies \cfrac{-1}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{-\sqrt{35}}{(\sqrt{35})^2}\implies -\cfrac{\sqrt{35}}{35}&#10;\\\\\\&#10;sec(\theta)=\cfrac{6}{\sqrt{35}}\implies \cfrac{6}{\sqrt{35}}\cdot \cfrac{\sqrt{35}}{\sqrt{35}}\implies \cfrac{6\sqrt{35}}{(\sqrt{35})^2}\implies \cfrac{6\sqrt{35}}{35}
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