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determine the minimum angle of elevation of H from John when he was walking along the jogging path ADE.

Guest Feb 14, 2015

Best Answer 

 #8
avatar+26406 
+5

Here's an approximate geogebra confirmation of Melody's result.

Scale everything so that h = 1.  Look down from above to see the following:

Walking route

The distances DE and AE are just 3/4 of distance AD (KD is √3).

The radius of the circle centred on K and going through E is approximately 2.04, so the angle of elevation from E to the top of the post is tan-1(1/2.04)

 

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{2.04}}}}\right)} = {\mathtt{26.113\: \!912\: \!630\: \!291^{\circ}}}$$

or approximately 26°

.

Alan  Feb 14, 2015
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9+0 Answers

 #1
avatar+91471 
+5

a)

AK=h     (right isosceles triangle AKH)

 

$$\\tan30=\frac{HK}{KD}\\\\
\frac{1}{\sqrt3}=\frac{h}{KD}\\\\
KD=\sqrt3\;h\\\\\\
AH^2+KD^2=AD^2\\
h^2+(\sqrt3\;h)^2=AD^2\\
AD^2=h^2+3h^2\\
AD^2=4h^2\\
AD=2h$$

 

b)

It takes 20 min to walk AD

15min to walk DE

and 15min to walk AE

so triangle ADE is isoceles and the ratio of the sides is 4:3:3

AD is 2h  so   DE=AE= (2h/4)*3 = 1.5h

Usine cosine rule 

 

$$\\cos(E)=\frac{3^2+3^2-4^2}{2*3*3}\\\\
cos(E)=\frac{9+9-16}{18}\\\\
cos(E)=\frac{1}{9}\\$$

 

$$\\$Area of \triangle ADE = $0.5*AE*ED*sinE$$

 

$$\\6272\sqrt5=0.5*1.5h*1.5h*sin(cos^{-1}\;\frac{1}{9}\;)\\\\
6272\sqrt5=0.5*1.5h*1.5h*\frac{\sqrt{80}}{9}\\\\
6272\sqrt5=\frac{1}{2}*\frac{3h}{2}*\frac{3h}{2}*\frac{4\sqrt{5}}
{9}\\\\
6272\sqrt5=\frac{9h^2}{8}*\frac{4\sqrt{5}}{9}\\\\
6272=\frac{h^2}{2}\\\\
h^2=12544\\\\
h=112$$

 

I simultaneously worked the problem and wrote the code so there could easily be error.

h did turn out to be a 'nice' number, so maybe it is right :)))

Feel free to ask questions 

 

I just realized that it is not finished yet.

I think the minumum angle of elevation TO H   is from E because that point is furthest away

I need to find distance EK

$$\\KD=\sqrt3*h\\
AE=ED=1.5h$$

 

This is not just falling out and I don't have more time now.  I shall return if another doesn't answer first. 

Melody  Feb 14, 2015
 #2
avatar+81051 
0

...................................

CPhill  Feb 14, 2015
 #3
avatar+81051 
+5

 

 

Melody is correct.....the minimum angle of elevation, from John's perespective, will occur at E, since this is the greatest distance from "h."

To find this angle of elevation, we need to know what EK is

Melody has already calculated angle E = cos-1 (1/9) = about 83.62°

And since triangle AED is isosceles, angle DAE = (180 - 83.62)/2 = 48.19°

And angle KAD can be found using the Law of Sines, thusly

sin KAD/KD = sin90/AD    and KD = 112/tan30  and AD =2h = 224

So we have

sin KAD =(112/tan30) /224   so sin-1 ((112/tan30)/224) = 60°

So angle DAE + KAD = angle KAE = 48.19 + 60 = 108.19°

And using the Law of Cosines, we have

EK^2 = AK^2 + AE^2 -2(AK)(AE)cos(108.19)

EK^2 = (112/tan30)^2 + 168^2 - 2(112/tan30)(168)cos(108.19)

EK = about 293.6

And the angle of elevation from E is given by

tan-1(112/293.6) = about 20.88°

 

Edit ....I have given the wrong measure for KA...it should be 112/tan 45 = 112

So we have

EK^2 = (112)^2 + 168^2 - 2(112)(168)cos(108.19)

EK = 229.162

And now my angle of elevation will = Melody's answer...!!!!  {more or less}

CPhill  Feb 14, 2015
 #4
avatar+91471 
+5

I wanted to do this answer exact but it is getting all too difficult and I don't know if what I have done is correct anyway. 

 

$$\\ED=1.5h=1.5*112\approx 168\\
KD=\sqrt3*h = \sqrt3*112\approx 194\\
\triangle AKD \equiv \triangle HKD \;\;\; $Three equal side test$\\\\
so\;\; KDA=30^0\\\\
But $Now consider triangle EKD$\\
EK^2=168^2+194^2-2*168*194*cos(78.2)$$

 

$${\sqrt{{{\mathtt{168}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{194}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{168}}{\mathtt{\,\times\,}}{\mathtt{194}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{cos}}{\left({\mathtt{78.2}}^\circ\right)}}} = {\mathtt{229.194\: \!522\: \!964\: \!077\: \!551\: \!3}}$$

$$\\EK \approx 229.2\\\\
Let \;\;\alpha $ be the angle of elevation sort, that is, $ tan\alpha = \frac{HK}{EK}\\\\
tan\alpha = \frac{h}{229.2}\\\\
tan\alpha = \frac{112}{229.2}\\\\
\alpha=tan^{-1}\frac{112}{229.2}\\\\$$

 

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{112}}}{{\mathtt{229.2}}}}\right)} = {\mathtt{26.042\: \!733\: \!855\: \!708^{\circ}}}$$

 

So if i have not made any stupid mistakes (which I probably did)  then  the minimum angle of elevation is approx 26 degrees and it is from the point E

Melody  Feb 14, 2015
 #5
avatar+91471 
0

Hey Chris, how come you got 3 points for an empty answer??   LOL

I want some of those too.    

Melody  Feb 14, 2015
 #6
avatar+81051 
0

HHAHA!!!...I wasn't reading the question that was presented and I started answering part A..I'll deduct that 3....(maybe!!!)

 

CPhill  Feb 14, 2015
 #7
avatar
+5

Melody , thank you for your help!!!!!!!!

Guest Feb 14, 2015
 #8
avatar+26406 
+5
Best Answer

Here's an approximate geogebra confirmation of Melody's result.

Scale everything so that h = 1.  Look down from above to see the following:

Walking route

The distances DE and AE are just 3/4 of distance AD (KD is √3).

The radius of the circle centred on K and going through E is approximately 2.04, so the angle of elevation from E to the top of the post is tan-1(1/2.04)

 

$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{1}}}{{\mathtt{2.04}}}}\right)} = {\mathtt{26.113\: \!912\: \!630\: \!291^{\circ}}}$$

or approximately 26°

.

Alan  Feb 14, 2015
 #9
avatar+91471 
0

Thanks Alan, the pic should help a lot :))

Melody  Feb 14, 2015

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