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4vir4ik [10]
3 years ago
8

JL is a common tangent to circles M and K at point J. If angle MLK measures 61ᵒ, what is the length of radius MJ? Round to the n

earest hundredth.

Mathematics
2 answers:
Sladkaya [172]3 years ago
7 0
3.122

The process to getting this comes in many steps. Firstly, you need to find the angles for JLK and MLJ. To find JLK use the arcsin function using the opposite side and the hypotenuse. 

Arcsin(Opp/Hype) = JLK
Arcsin(.5) = JLK
30 degrees = JLK

This means MLJ = 31 degrees since they add up to 61 degrees. 

Now we need to find the length of LJ, which we can do using the Pythagorean Theorem. 

3^2 + JL^2 = 6^2
9 + JL^2 = 36
JL^2 = 27
JL = \sqrt{27}

Now that we have the angle of MLJ and the length of JL, we can use the tangent function to find MJ.

Tan(angle) = opp/adj
Tan(31) = MJ/\sqrt{27}
\sqrt{27}Tan(31) = MJ
3.122 = MJ
laila [671]3 years ago
6 0
ΔJKL, sin(JLK)=JK/KL=3/6=0.5
<JKL=30deg.

cos(JLK)=JL/KL
JL=KL*cos(JLK)=6cos(30deg.)

<JLK + <JLM = <MLK
30 + <JLM = 61
<JLM=31deg

in ΔMJL
tan(JLM)=JM/JL
JM=JL*tan(JLM)
=6cos(30deg.)*tan(31deg)
=3.12
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A rectangular parking lot has an area of 15,000 feet squared, the length is 20 feet more than the width. Find the dimensions
faust18 [17]

Dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet

<h3><u>Solution:</u></h3>

Given that  

Area of rectangular parking lot = 15000 square feet

Length is 20 feet more than the width.

Need to find the dimensions of rectangular parking lot.

Let assume width of the rectangular parking lot in feet be represented by variable "x"

As Length is 20 feet more than the width,

so length of rectangular parking plot = 20 + width of the rectangular parking plot

=> length of rectangular parking plot = 20 + x = x + 20

<em><u>The area of rectangle is given as:</u></em>

\text {Area of rectangle }=length \times width

Area of rectangular parking lot = length of rectangular parking plot \times width of the rectangular parking

\begin{array}{l}{=(x+20) \times (x)} \\\\ {\Rightarrow \text { Area of rectangular parking lot }=x^{2}+20 x}\end{array}

But it is given that Area of rectangular parking lot = 15000 square feet

\begin{array}{l}{=>x^{2}+20 x=15000} \\\\ {=>x^{2}+20 x-15000=0}\end{array}

Solving the above quadratic equation using quadratic formula

<em><u>General form of quadratic equation is  </u></em>

{ax^{2}+\mathrm{b} x+\mathrm{c}=0

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x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}

In our case b = 20, a = 1 and c = -15000

Calculating roots of the equation we get

\begin{array}{l}{x=\frac{-(20) \pm \sqrt{(20)^{2}-4(1)(-15000)}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{400+60000}}{2 \times 1}} \\\\ {x=\frac{-(20) \pm \sqrt{60400}}{2}} \\\\ {x=\frac{-(20) \pm 245.764}{2 \times 1}}\end{array}

\begin{array}{l}{=>x=\frac{-(20)+245.764}{2 \times 1} \text { or } x=\frac{-(20)-245.764}{2 \times 1}} \\\\ {=>x=\frac{225.764}{2} \text { or } x=\frac{-265.764}{2}} \\\\ {=>x=112.882 \text { or } x=-132.882}\end{array}

As variable x represents width of the rectangular parking lot, it cannot be negative.

=> Width of the rectangular parking lot "x" = 112.882 feet  

=> Length of the rectangular parking lot = x + 20 = 112.882 + 20 = 132.882

Hence can conclude that dimension of rectangular parking lot is width = 112.882 feet and length = 132.882 feet.

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