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November 3rd, 2016, 07:33 AM   #1
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Height and distance

From top of a tower the angle of depression of the top and bottom of a multistoreyed building are 30° and 60° respectively. If the height of the building is 100m, find the height of the tower.

Last edited by skipjack; November 3rd, 2016 at 07:42 AM.
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November 3rd, 2016, 08:13 AM   #2
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see attached sketch

h = tower height

x = horizontal distance between tower and building

solve the system of equations for h ...
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File Type: jpg tower height.jpg (32.3 KB, 12 views)
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November 3rd, 2016, 08:30 AM   #3
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By the angle bisector theorem, (h - 100m)/100m = x/h = cos(60°) = 1/2, where x and h are as in skeeter's diagram, so h = 150m.
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November 3rd, 2016, 09:44 AM   #4
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A triangle is 30-60-90, $(2x)^2=x^2+h^2\implies3x^2=h^2\implies x=\frac{h}{\sqrt3}$.

Substitute for $x$ in the equations to the right of the diagram and subtract:

$$100=\frac{h}{\sqrt3}(\tan60-\tan30)=h\left(1-\frac13\right)\implies h=150\text{ m}$$
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November 3rd, 2016, 12:26 PM   #5
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The 30°-30°-120° triangle has shorter sides both of length 100m.

As the small 30°-60°-90° triangle above it is half of an equilateral triangle,
h - 100m = (100m)/2, and so h = 150m.
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November 4th, 2016, 05:00 PM   #6
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Quote:
Originally Posted by skipjack View Post
By the angle bisector theorem, (h - 100m)/100m = x/h = cos(60°) = 1/2, where x and h are as in skeeter's diagram, so h = 150m.
totally forgot about the angle bisector theorem.
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November 4th, 2016, 10:50 PM   #7
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November 5th, 2016, 03:39 AM   #8
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Using $3x^2=h^2$ and similar triangles (from skeeter's diagram),

$$\frac{h-100}{x}=\frac xh$$

$$h^2-100h=x^2$$

$$3h^2-300h=3x^2=h^2\implies h=150$$
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