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# Urgent help needed!!!! but i want to know how to solve this myself(no cheating)!

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Joe places a mirror on the ground 60ft from the base of his school's clock tower. He walks backwards untill he can see the top of the tower in the mirror. At that point, his eyes are 6 ft above the ground, and he is 7.5 ft away from the mirror. How tall is the clock tower? How do I even begin to solve this????

JessicaAishaRenee  Feb 17, 2016

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Joe places a mirror on the ground 60ft from the base of his school's clock tower. He walks backwards untill he can see the top of the tower in the mirror. At that point, his eyes are 6 ft above the ground, and he is 7.5 ft away from the mirror. How tall is the clock tower? How do I even begin to solve this????

The important thing to know here is that the angle of incidence must equal the angle of reflection.

So now this is our situation

Angle A and A' are both theta degres because they are altenate angles on parallel lines with th original thetas (angle of incident and angle of reflections.

This it is evident that these 2 triangles are similar,

so just like Dragonlance said.

$$\frac{h}{60}=\frac{6}{7.5}\\ h=\frac{6}{7.5}*60\\$$

6/7.5*60 = 48 feet

Melody  Feb 17, 2016
edited by Melody  Feb 17, 2016
edited by Melody  Feb 17, 2016
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No idea- sorry

Mathfordays  Feb 17, 2016
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thx anyways!

JessicaAishaRenee  Feb 17, 2016
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So the mirror is 60ft from the base of his school clock tower b

He walks backwards an x amount of times until he can see the top of the clock tower in the mirror,

At this point, his eyes are 6 ft above ground, and he's 7.5 ft away from the mirror, .

Let the top of the tower be T and let M be mirror

So we know the tower is $$T\geq 6$$

He walked 7.5ft away from the mirror so let x=7.5ft

So we now have the base lenght and height I.e 2 sides

so we know that the max height of the Tower cannot excead 60ft since he would be going past his mirror.

So $$T\leq 60$$  so now we have this inequality

$$6\leq T\leq 60$$

This is actually a Pythagores question since we know the max heght can't excead 60 and can't go below 6

so it's 60^2-7.5^2=c^2

So the height is 59.6 to 1 decimal place! *Made a mistake instead of 6 we are subtracting 7.5 ft so

its 60^2-7.5^2=3543 and the sqrt of that is 59.5 to 1 decimal place! sorry for the mistake

Misaki  Feb 17, 2016
edited by Misaki  Feb 17, 2016
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Height of Joe’s eyes divided by the distance from the mirror equals the height of the tower divided by the distance to mirror.

H/D = h/d

H/60 = 6/7.5

H= (6/7.5)*60

H= 48

The tower is 48 feet high.

Dragonlance  Feb 17, 2016
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Dang, I come on right as you leave :(

Coldplay  Feb 17, 2016
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Very nice, Dragonlace.....I would have solved it using trig, but your approach is much simpler....!!!!

CPhill  Feb 17, 2016
#7
+92437
+5

Joe places a mirror on the ground 60ft from the base of his school's clock tower. He walks backwards untill he can see the top of the tower in the mirror. At that point, his eyes are 6 ft above the ground, and he is 7.5 ft away from the mirror. How tall is the clock tower? How do I even begin to solve this????

The important thing to know here is that the angle of incidence must equal the angle of reflection.

So now this is our situation

Angle A and A' are both theta degres because they are altenate angles on parallel lines with th original thetas (angle of incident and angle of reflections.

This it is evident that these 2 triangles are similar,

so just like Dragonlance said.

$$\frac{h}{60}=\frac{6}{7.5}\\ h=\frac{6}{7.5}*60\\$$

6/7.5*60 = 48 feet

Melody  Feb 17, 2016
edited by Melody  Feb 17, 2016
edited by Melody  Feb 17, 2016