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aivan3 [116]
3 years ago
11

A 20 ft. ladder is used against a 15 ft. wall. What is the measure of the angle made by the ladder and the ground (nearest whole

degree)? How far is the ladder from the wall on the ground (to the nearest tenth)?
Mathematics
1 answer:
Brrunno [24]3 years ago
6 0

Step-by-step explanation:

Given that,

The length of a ladder, H = 20 feet

The height of the wall, h = 15 ft

We know that,

\sin\theta=\dfrac{h}{H}

h is perpendicular and H is hypotenuse

So,

\sin\theta=\dfrac{15}{20}\\\\\theta=\sin^{-1}(\dfrac{15}{20})\\\\\theta=48.59^{\circ}

Now using Pythagoras theoerm,

b=\sqrt{H^2-h^2}\\\\b=\sqrt{20^2-15^2}\\\\b=13.2\ ft

Hence, the angle made by the ladder and the ground is 48.59° and the ladder is 13.2 feet from the wall on the ground.

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3 years ago
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katen-ka-za [31]

Question # 13

Answer:

The required equation for the given function is <em>y = 4sin(x/2+2π/3) -2 , as shown attached graph diagram.</em>

<em>Step-by-step explanation: </em>

As the general sine function is given by

y=asin(bx+c)+d.......[A]

  • amplitude = a
  • period = 2π ÷ b
  • Phase shift = -c ÷ b
  • Vertical shift = d

As in the question,  

  • amplitude = a = 4
  • period = 4π
  • phase shift = -4π/3
  • Vertical shift = d = -2

As  

period = 2π ÷ b  

b = 2π/period

b = 2π/4π ∵ period = 4π

b = 1/2  

Also

Phase shift = -c/b

-4π/3 = -c/b ∵ phase shift = -4π/3

4π/3 = c/b  

c = b × 4π/3  

c = 1/2 × 4π/3  

c = 4π/6  

c = 2π/3

So, putting Amplitude ⇒ a = 4, Vertical shift ⇒ d = -2, b = 1/2 ,  

and c = 2π/3 in Equation [A] would bring us the required equation for the given function.

y=asin(bx+c)+d

y = 4sin(x/2+2π/3)+(-2)

y = 4sin(x/2+2π/3) -2            

<em>Note: The graph is also shown in attached diagram.</em>

                                             Question # 14

<em>Answer:</em>

The required equation for the given function is y = cot(x+π/3)+2, as shown in attached graph diagram.

<em>Step-by-step explanation: </em>

As the general cotangent function is given by

y=acot(bx+c)+d.......[A]

  • amplitude = a
  • period = π ÷ b
  • Phase shift = -c ÷ b
  • Vertical shift = d

As in the question,  

  • period = π
  • phase shift = -π/3
  • Vertical shift = d = 2

As  

period = π ÷ b  

b = π/period

b = π/π ∵ period = 4π

b = 1  

Also

Phase shift = -c/b

-π/3 = -c/b ∵ phase shift = -π/3

π/3 = c/b  

c = b × π/3  

c = 1 × π/3  

c = π/3

So, putting vertical shift ⇒ d = 2, b = 1 and   c = π/3 in Equation [A] would bring us the required equation for the given function.

y=acot(bx+c)+d

y = cot(x+π/3)+2

<em>Note: The graph is also shown in attached diagram.</em>

Keywords: amplitude, period , phase shift , vertical shift

Learn more about trigonometric functions of equations from brainly.com/question/2643311

#learnwithBrainly

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Answer:

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Step-by-step explanation:

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