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3 years ago

Determine the sign of cos pi divided by seven without using a calculator.

Mathematics

2 answers:

Sophie [7]3 years ago

3 0

The angle $\frac{\pi}{7}$ is in Quadrant 1 and cos of an angle in quadrant 1 is always positive

Aliun [14]3 years ago

3 0

Answer:

The $\cos \frac{\pi}{7}$ has positive sign.

Step-by-step explanation:

we need to determine the sign of $\cos \frac{\pi}{7}$ without using calculator

Cartesian system ( shown in figure -1)

The angle $\cos \frac{\pi}{7}$ lies in quadrant-1 , Where angles are zero (0) to $\frac{\pi}{2}$

and in first quadrant all trigonometric ratios has positive sign

so, $\cos \frac{\pi}{7}$ has positive sign.

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kow [346]

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2 years ago

A 45 foot ladders leaning up against a building. The base of the latter is 20 feet away from the building. What is the angle of

Tatiana [17]

Answer:

The angle of elevation to the top of the building is 63.61 degrees

Step-by-step explanation:

Here, we want to calculate angle of elevation to the top of the building.

For this, we need a triangle

Please check for this in the attachment.

From the diagram, we are to calculate the angle theta.

To do this, we use trigonometric identities.

Looking at what we have, we have the hypotenuse and the adjacent.

So the trigonometric identity to use is the cosine

Mathematically Cosine theta = adjacent/hypotenuse

Thus, Cos theta = 20/45

Cos theta = 0.444444444444444

Theta = Arc cos(0.444444444444444)

Theta = 63.61 degrees

8 0

3 years ago

Find the slope between the pair of points (1,3) and (7,2)

agasfer [191]

Answer:

$\bold{m=-\dfrac16}$

Step-by-step explanation:

$\bold{slope\, (m)=\dfrac{change\ in\ Y}{change\ in\ X}=\dfrac{y_2-y_1}{x_2-x_1}}$

(1, 3) ⇒ x₁ = 1, y₁ = 3

(7, 2) ⇒ x₂ = 7, y₂ = 2

So the slope:

$\bold{m=\dfrac{2-3}{7-1}=\dfrac{-1}6=-\dfrac16}$

5 0

3 years ago

Question number 8. Prove that 3 + √5 is an irrational number.

Marina86 [1]

Answer :

Let's assume the opposite of the statement i.e., 3 + √5 is a rational number.

$\\ \sf \: 3 + \sqrt{5} = \frac{a}{b} \\ \\ \qquad \: \tiny \sf{(where \: \: a \: \: and \: \: b \: \: are \: \: integers \: \: and \: \: b \: \neq \: 0)} \\$

$\\ \sf \: \sqrt{5} = \frac{a}{b} - 3 = \frac{a - 3b}{b} \\$

Since, a, b and 3 are integers. So,

$\\ \sf \: \frac{p - 3b}{b} \\ \\ \qquad \tiny \sf{ \: (is \: \: a \: \: rational \: \: number \:) } \\$

Here, it contradicts that √5 is an irrational number.

because of the wrong assumption that 3 + √5 is a rational number.

$\\$

Hence, 3 + √5 is an irrational number.

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3 years ago

Anja simplified the expression StartFraction 10 x Superscript negative 5 Baseline Over Negative 5 x Superscript 10 Baseline EndF

katovenus [111]

Answer:

She subtracted the coefficients instead of dividing them

Step-by-step explanation:

7 0

3 years ago

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