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charle [14.2K]
3 years ago
12

The side of a hill makes a 12° with the horizontal. A wire is to be run from the top of a 175 foot tower on the top of the hill

to a stake 120 ft down the hillside from the base of the tower. What length of wire is needed
Mathematics
1 answer:
o-na [289]3 years ago
8 0

Answer:

231.9ft

Step-by-step explanation:

We would be solving this problem by using the cosine rule.

The length = (pq)² + (qr)² -2 * pq * qr * cos102⁰

= 120² + 175² - 2*120*175 (-0.2079)

= 14400 + 30625 + 8731.8

= 53756.8

Length² = 53756.8

We take the square root of both sides

Length = √53756.8

Length = 231.89 feet

This is the length of the wire that is needed.

Thank you

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A city planner designs a park that is a quadrilateral with vertices at J(-3,1), K(1,3), L(5,-1), and M(-1,-3). There is an entra
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The Quadrilateral is JKLM, 

let M_{JK}, M_{KL}, M_{LM}, M_{JM}, be the midpoints of JK, KL, LM and JM respectively.

---------------------------------------------------------------------------------------------------------

Given any 2 point P(m,n) and Q(k,l),<span>

the coordinates of the midpoint of the line segment PQ are given by the formula:

M_{PQ}=( \frac{m+k}{2} ,&#10;\frac{n+l}{2}), </span>

-------------------------------------------------------------------------------------------------

thus the coordinates of points M_{JK}, M_{KL}, M_{LM}, M_{JM},

are as follows:

M_{JK}= (\frac{-3+1}{2}, \frac{1+3}{2})=(-1,2), \\\\M_{KL}= (\frac{1+5}{2}, \frac{3-1}{2}=(3,1), \\\\M_{LM}= (\frac{5-1}{2}, \frac{-1-3}{2})=(2, -2),\\\\ M_{JM}= (\frac{-3-1}{2}, \frac{1-3}{2})=(-2,-1)


------------------------------------------------------------------------------------------------

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:<span>

 |PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2}&#10;}</span>

------------------------------------------------------------------------------------------------

thus the distances connecting the opposite entrances can be calculated as follows:


|M_{JK},M_{LM}|= \sqrt{ (-1-2)^{2} + (2-(-2))^{2} }= \sqrt{9+16}=5

|M_{KL}M_{JM}|= \sqrt{ (3-(-2))^{2} + (1-(-1))^{2}}= \sqrt{25+4}= \sqrt{29}=5.39


Thus the total distance of the paths joining the opposite entrances is 

5+5.39 units = 50 m + 53.9 m = 104 m (rounded to the nearest meter)


Answer: 104 m



8 0
3 years ago
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