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3 years ago

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Triangle ABC is a right triangle and sin(53o) = StartFraction 4 Over x EndFraction. Solve for x and round to the nearest whole n

umber.
Triangle A B C is shown. Angle A C B is a right angle and angle B A C is 53 degrees. The length of B C is 4 centimeters, the length of A C is y, and the length of hypotenuse A B is x.

Which equation correctly uses the value of x to represent the cosine of angle A?

cos(53o) = StartFraction 4 Over x EndFraction
cos(53o) = StartFraction y Over 5 EndFraction
cos(53o) = StartFraction x Over 4 EndFraction
cos(53o) = StartFraction 5 Over y EndFraction

1 answer:

zmey [24]3 years ago

4 0

Answer:

cos(53°)=StartFraction y Over 5 EndFraction

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the value of x

we know that

$sin(53\°)=\frac{4}{x}$

Solve for x

$x=\frac{4}{sin(53\°)}$

$x=5$

step 2

Find the cosine of angle A

The cosine of angle A is equal to divide the adjacent side to angle A (AC) by the hypotenuse (AB)

$cos(53\°)=\frac{y}{x}$

substitute the value of x

$cos(53\°)=\frac{y}{5}$

so

cos(53°)=StartFraction y Over 5 EndFraction

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If you are offered one slice from a round pizza (in other words, a sector of a circle) and the slice must have a perimeter of 24

VARVARA [1.3K]

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A 12 in diameter will reward you with the largest slice of pizza.

Step-by-step explanation:

Let $r$ be the radius and $\theta$ be the angle of a circle.

According with the graph, the area of the sector is given by

$A=\frac{1}{2}r^2\theta$

The arc lenght of a circle with radius $r$ and angle $\theta$ is $r \theta$

The perimeter of the pizza slice is composed of two straight pieces, each of length r inches, and an arc of the circle which you know has length s = rθ inches. Thus the perimeter has length

$2r+r\theta=24 \:in$

We need to express the area as a function of one variable, to do this we use the above equation and we solve for $\theta$

$2r+r\theta=24\\r\theta=24-2r\\\theta=\frac{24-2r}{r}$

Next, we substitute this equation into the area equation

$A=\frac{1}{2}r^2(\frac{24-2r}{r})\\A=\frac{1}{2}r(24-2r)\\A=12r-r^2$

The domain of the area is

$0$

To find the diameter of pizza that will reward you with the largest slice you need to find the derivative of the area and set it equal to zero to find the critical points.

$\frac{d}{dr} A=\frac{d}{dr}(12r-r^2)\\A'(r)=\frac{d}{dr}(12r)-\frac{d}{dr}(r^2)\\A'(r)=12-2r$

$12-2r=0\\-2r=-12\\\frac{-2r}{-2}=\frac{-12}{-2}\\r=6$

To check if $r=6$ is a maximum we use the Second Derivative test

if $f'(c)=0$ and $f''(c)$ , then f(x) has a local maximum at x = c.

The second derivative is

$\frac{d}{dr} A'(r)=\frac{d}{dr} (12-2r)\\A''(r)=-2$

Because $A''(r)=-2$ the largest slice is when r = 6 in.

The diameter of the pizza is given by

$D=2r=2\cdot 6=12 \:in$

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3 years ago

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masya89 [10]

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Step-by-step explanation:

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Step-by-step explanation:

From the question we are told that

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