︎help me please ineed answer asp thankyou

Mathematics

1 answer:

Arisa [49]2 years ago

5 0

Answer:

Look below

Step-by-step explanation:

I'm going to use B as my example to explain: B (-3,4)

You are going to need to draw a coordinate plane. When drawing the X axis, you want the highest X point to be 6 due to the fact that that is your highest x value. Then you want the lowest X point to be -6. Same will apply with the Y axis.

Next you are going to put a point in the middle of the plane which is called the origin (0,0).

Going from the origin, you want to follow the x axis the get to -3 which will be going left since its negative. Then since the Y axis is vertical, you are going to go up 4 times since the 4 is positive.

Then start from the origin for the next point being C (-5,4), Remember to always start from the origin.

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Now, the cosecant of θ is -6, or namely -6/1.

however, the cosecant is really the hypotenuse/opposite, but the hypotenuse is never negative, since is just a distance unit from the center of the circle, so in the fraction -6/1, the negative must be the 1, or 6/-1 then.

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\\\\\\
\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}
\\\\\\
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recall that

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

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

$\bf sin(\theta)=\cfrac{-1}{6}
\qquad\qquad 
cos(\theta)=\cfrac{\sqrt{35}}{6}
\\\\\\
% tangent
tan(\theta)=\cfrac{-1}{\sqrt{35}}
\qquad \qquad 
% cotangent
cot(\theta)=\cfrac{\sqrt{35}}{1}
\\\\\\
sec(\theta)=\cfrac{6}{\sqrt{35}}$

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\\\\\\
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