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Bumek [7]
3 years ago
13

Ship A and Ship B are 120 km apart when they pick up a distress call from another boat. Ship B estimates that they are 70 km awa

y from the distress call. They also notice that the angle between the line from ship B to ship A and the line from ship A to the distress call is 28°. What are the two possible distances, to the nearest TENTH of a km, from ship A to the boat?

Mathematics
1 answer:
Greeley [361]3 years ago
5 0

Answer:

The two possible distance are 147.5km and 64.4km

Step-by-step explanation:

Given

<em>See Attachment for Illustration</em>

Required

Determine the possible distance of ship A from the boat

The distress is represented by X;

So, the question requires we calculate distance AX;

This will be done using cosine formula as follows;

a^2 = b^2 + x^2 - 2bxCosA

In this case;

a = 70;

b = ??

x = 120

A = 28 degrees

Substitute these values in the formula above

70^2 = b^2 + 120^2 - 2 * b * 120 Cos28

4900 = b^2 + 14400 - 2 * b * 120 * 0.8829

4900 = b^2 + 14400 - 211.9b

Subtract 4900 from both sides

b^2 + 14400 - 4900 - 211.9b = 0

b^2 + 9500 - 211.9b = 0

b^2 - 211.9b + 9500  = 0

Solve using quadratic formula

\frac{-b\±\sqrt{b^2 - 4ac}}{2a}

Substitute 1 for a; -211.9 for b and 9500 for c

b = \frac{-(211.9)\±\sqrt{(211.9)^2 - 4 * 1 * 9500}}{2 * 1}

b = \frac{211.9\±\sqrt{44901.61 - 38000}}{2}

b = \frac{211.9\±\sqrt{6901.61}}{2}

b = \frac{211.9\±83.08}{2}

This can be splitted to

b = \frac{211.9+83.08}{2}  or    b = \frac{211.9-83.08}{2}

b = \frac{294.98}{2}   or  b = \frac{128.82}{2}

b = 147.49   or   b = 64.41

b = 147.5km\ or\ b = 64.4km

<em>Hence, the two possible distance are 147.5km and 64.4km</em>

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Step-by-step explanation:

Please find the attachment.

We have been given that a norman window is constructed by adjoining a semicircle to the top of an ordinary rectangular. The total perimeter is 38 feet.

The perimeter of the window will be equal to three sides of rectangle plus half the perimeter of circle. We can represent our given information in an equation as:

2L+W+\frac{1}{2}(2\pi r)=38

We can see that diameter of semicircle is W. We know that diameter is twice the radius, so we will get:

2L+W+\frac{1}{2}(2r\pi)=38

2L+W+\frac{\pi}{2}W=38

Let us find area of window equation as:

\text{Area}=W\cdot L+\frac{1}{2}(\pi r^2)

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\text{Area}=W\cdot L+\frac{\pi}{8}W^2

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Substitute this value in area equation:

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Since we need the area of window to maximize, so we need to optimize area equation.

A=W\cdot (38-W-\frac{\pi }{2}W)+\frac{\pi}{8}W^2  

A=38W-W^2-\frac{\pi }{2}W^2+\frac{\pi}{8}W^2  

Let us find derivative of area equation as:

A'=38-2W-\frac{2\pi }{2}W+\frac{2\pi}{8}W  

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A'=38-2W-\frac{4\pi W}{4}+\frac{\pi}{4}W

A'=38-2W-\frac{3\pi W}{4}

To find maxima, we will equate first derivative equal to 0 as:

38-2W-\frac{3\pi W}{4}=0

-2W-\frac{3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}=-38

\frac{-8W-3\pi W}{4}*4=-38*4

-8W-3\pi W=-152

8W+3\pi W=152

W(8+3\pi)=152

W=\frac{152}{8+3\pi}

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Upon substituting W=8.723210 in equation L=38-(W+\frac{\pi }{2}W), we will get:

L=38-(8.723210+\frac{\pi }{2}8.723210)

L=38-(8.723210+\frac{8.723210\pi }{2})

L=38-(8.723210+\frac{27.40477245}{2})

L=38-(8.723210+13.70238622)

L=38-(22.42559622)

L=15.57440378

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Therefore, the dimensions of the window that will maximize the area would be W\approx 8.72 and L\approx 15.57.

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