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ASHA 777 [7]
3 years ago
8

Find cos , csc 0, and tan 0, where is the angle shown in the figure.

Mathematics
1 answer:
docker41 [41]3 years ago
5 0

Answer:

Step-by-step explanation:

<u>Trigonometric Ratios</u>

The ratios of the sides of a right triangle are called trigonometric ratios. There are six trigonometric ratios, sine (sin), cosine (cos), tangent (tan), cosecant (csc), secant (sec), and cotangent (cot).

The ratios are valid only in right triangles, or when one of the internal angles is 90°.

Take any of the acute angles as a reference, and the largest side as the hypotenuse, then:

\displaystyle \cos\theta=\frac{\text{adjacent leg}}{\text{hypotenuse}}

\displaystyle csc\theta=\frac{\text{hypotenuse}}{\text{opposite leg}}

\displaystyle \tan\theta=\frac{\text{opposite leg}}{\text{adjacent leg}}

The missing side of the triangle x can be calculated by using Pythagora's Theorem:

10^2=9^2+x^2

Solving:

x^2=10^2-9^2=100-81=19

x=\sqrt{19}

For the angle marked as θ:

\displaystyle \tan\theta=\frac{\text{opposite leg}}{\text{adjacent leg}}

\displaystyle \tan\theta=\frac{\text{9}}{\sqrt{19}}

Rationalizing:

\displaystyle \tan\theta=\frac{9\sqrt{19}}{19}

\displaystyle \cos\theta=\frac{\text{adjacent leg}}{\text{hypotenuse}}

\displaystyle \cos\theta=\frac{\sqrt{19} }{10}

\displaystyle \csc\theta=\frac{\text{hypotenuse}}{\text{opposite leg}}

\displaystyle \csc\theta=\frac{10}{9}

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x ≈ 6

Step-by-step explanation:

The legs of an isosceles triangle are congruent , both 2x - 4 , then

x + 2x - 4 + 2x - 4 = 24 , that is

5x - 8 = 24 ( add 8 to both sides )

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What is a name for a marked angle ? Answers to choose from :
katen-ka-za [31]

Answer:

Angles are named in two ways. You can name a specific angle by using the vertex point, and a point on each of the angle's rays. The name of the angle is simply the three letters representing those points, with the vertex point listed in the middle. You can also name angles by looking at their size.

Step-by-step explanation:

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25 POINTS, I DONT UNDERSTANDDDDDD :C WHAT IS SO MISLEADING ABOUT THIS GRAPH???
Kisachek [45]

Answer:

honestly, the graph look totally fine...

If one ere pressed to find something to complain about it, one could suggest that  you do not know if this was the starting price of the stock or the ending price of the stock each day?... One could also argue that to be a bit more meaningful you might want to know the range of prices during each day...

look up what is called a candle stick graph.. each day looks like a candlestick... the top is the highest value each the bottom the lowest, and there is a line in the candle that  shows the closing price

Step-by-step explanation:

3 0
3 years ago
find the angle between the vectors. (first find the exact expression and then approximate to the nearest degree. ) a=[1,2,-2]. B
SashulF [63]

Answer:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

Step-by-step explanation:

For this case first we need to calculate the dot product of the vectors, and after this if the dot product is not equal to 0 we can calculate the angle between the two vectors in order to see if there are parallel or not.

a=[1,2,-2], b=[4,0,-3,]

The dot product on this case is:

a b= (1)*(4) + (2)*(0)+ (-2)*(-3)=10

Since the dot product is not equal to zero then the two vectors are not orthogonal.

Now we can calculate the magnitude of each vector like this:

|a|= \sqrt{(1)^2 +(2)^2 +(-2)^2}=\sqrt{9} =3

|b| =\sqrt{(4)^2 +(0)^2 +(-3)^2}=\sqrt{25}= 5

And finally we can calculate the angle between the vectors like this:

cos \theta = \frac{ab}{|a| |b|}

And the angle is given by:

\theta = cos^{-1} (\frac{ab}{|a| |b|})

If we replace we got:

\theta = cos^{-1} (\frac{10}{\sqrt{9} \sqrt{25}})=cos^{-1} (\frac{10}{15}) = cos^{-1} (\frac{2}{3}) = 48.190

Since the angle between the two vectors is not 180 or 0 degrees we can conclude that are not parallel

And the anfle is approximately \theta \approx 48

3 0
3 years ago
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