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Misha Larkins [42]

4 years ago

5

What are the formulas for COS, TAN, and SIN?

1 answer:

Doss [256]4 years ago

8 0

I'm guessing that you don't really want the formulas. I think what you actually want
is the definitions of those functions of an acute angle when it's in a right triangle.

Cosine = (adjacent side) / (hypotenuse)

Tangent = (opposite side) / (adjacent side)

Sine = (opposite side) / (hypotenuse)

Tell me if I'm wrong, and I'll find some formulas for you.

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A research center poll showed that 76% of people believe that it is morally wrong to not report all income on tax returns. What

erastovalidia [21]

Answer:

6/25

Step-by-step explanation:

P(have belief) =76/100 = 19/25

P (does not have belief) = 1-19/25 = 6/25

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3 years ago

Determine the cardinal number of A={2,4,6,8,10}

dimulka [17.4K]

A cardinal number of a set is the number of elemenents it contains. Therefore the cardinal number of A is equal to 5.

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3 years ago

Light travels at about 3000 000 000 meter per second what is this speed in kilometers per hour

faltersainse [42]

3000000 Kilometers i believe

6 0

3 years ago

Read 2 more answers

madam [21]

Answer:

D

Step-by-step explanation:

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4 0

3 years ago

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Please find the perimeter and area of these shapes

ladessa [460]

Answer:

ABC shaded area = 36 $\pi$ - 72 cm²

ABC shaded area perimeter = $6\pi +12\sqrt{2}$ cm

ABCD area = $\dfrac52 \pi$ cm²

ABCD perimeter = $3\pi +2$ cm

Step-by-step explanation:

<u>Shape ABC</u>

Assuming you want the area and perimeter of the shaded part of the shape only...

<u>Area</u>

Area of a sector = $\dfrac12r^2\theta$ (where r is the radius and $\theta$ <em> </em>

⇒ area of a sector = $\dfrac12 \times 12^2\times \dfrac{\pi}{2} =36\pi \ \textsf{cm}^2$

Area of triangle = 1/2 x base x height

⇒ area of triangle = 1/2 x 12 x 12 = 72 cm²

Therefore, area of shaded area = area of sector - area of triangle

⇒ area = 36 $\pi$ - 72 cm²

<u>Perimeter</u>

Arc length = $r\theta$ (where r is the radius and $\theta$ <em> </em>

⇒ arc length = $12\times\dfrac12\pi =6\pi \ \textsf{cm}$

Hypotenuse of triangle = $\sqrt{a^2+b^2}$ (where a and b are the legs of the right triangle)

⇒ hypotenuse = $\sqrt{12^2+12^2} =12\sqrt{2}$ cm

Therefore, perimeter = arc length + hypotenuse

⇒ perimeter = $6\pi +12\sqrt{2}$ cm

<u>Shape ABCD</u>

<u>Area</u>

Area of a semicircle = $\dfrac12 \pi r^2$ (where r is the radius)

⇒ area of large semicircle ABC = $\dfrac12 \times \pi \times 2^2=2\pi \ \textsf{cm}^2$

⇒ area of small semicircle AD = $\dfrac12 \times \pi \times 1^2=\dfrac12\pi \ \textsf{cm}^2$

⇒ area of shape ABCD = $\dfrac12 \pi + 2 \pi=\dfrac52 \pi \ \textsf{cm}^2$

<u>Perimeter</u>

1/2 circumference = $\pi r$

⇒ perimeter = $2\pi +2+\pi=3 \pi+2 \ \textsf{cm}$

7 0

2 years ago