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daser333 [38]
3 years ago
9

Find the exact value of cos(sin^-1(-5/13))

Mathematics
1 answer:
son4ous [18]3 years ago
8 0

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}}

\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}

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.

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How would you set up this problem to find the height of the sail using reciprocal functions?
AVprozaik [17]

Remark

I'm not sure I know what you mean, but I will answer trying to interpret it so it fits the description of the question.

Step One: Set up the equation. The normal way of solving this would be to use the Cos function.

Cos(theta) = Adjacent / hypotenuse = x/8 Multiply both sides by 8

8*cos(theta) = x Theta = 35

8*cos(35) = x This is about as simple a way as is possible.

6.553 = the height of the sale.

Step Three: try using the reciprocal

1 / cos(35)= hypotenuse / adjacent

1/0.8191 = 8/x cross multiply

x = 0.8191 * 8

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5 0
3 years ago
A line and a circle intersect at the points A and B. Use the equations below to find the coordinates of the points of intersecti
Margaret [11]

Answer:

<h2>A(-2, 2) and B(6, 10)</h2>

Step-by-step explanation:

Given the equation of a line y = x + 4 and equation of a circle as

( x − 3 )² + ( y − 5 )² = 34, if the line and the circle intersect at points A and B, to get this points, we will substitute the equation of the line into that of the circle as shown;

We will have to expand the equation of the circle first before making the substitute.

( x − 3 )² + ( y − 5 )² = 34

x²-6x+9+y²-10y+25 = 34

x²+y²-6x+-10y+34-34 = 0

x²+y²-6x+-10y = 0

Substituting y = x+ 4 into the resulting expression;

x²+(x+4)²-6x+-10y = 0

x²+x²+8x+16-6x+-10(x+4) = 0

x²+x²+8x+16-6x+-10x-40 = 0

2x²-8x-24 = 0

x²-4x-12 = 0

(x²-6x)+(2x-12) = 0

x(x-6)+2(x-6) = 0

x+2 = 0 and x-6 = 0

x = -2 and 6

when x = -2;

y = -2+4

y = 2

when x = 6

y = 6+4

y = 10

The coordinates of the point of intersection are A(-2, 2) and B(6, 10).

4 0
3 years ago
Jadwiga runs for 875 meters everyday. After 7 days, how many kilometers has Jadwiga run
OverLord2011 [107]

Given data:

The given distance Jadwiga runs in a day is 875 m.

The expression for the given statement is,

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Multiply the above expression by 7 on both sides.

7(1 day)=7(875 m)

=6,125 m

=6.125 km

Thus, Jadwiga covered 6.125 km in 7 days.

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