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HOW TALL IS IT?
Taylor SchmelzleA.J. LeatherwoodValerie BarreauAdam Cooper
6th Period 9 March 2009
3o degrees Taylor Calculations: Long Leg = √ 3 × short leg
28 = √3 × t
28 / √3 = t
≈ 16.17 ft + 5 ft
≈ 21.17 ft Tangent t = opposite /
adjacent
tan 30= t / 28
28 (tan 30) = t
t ≈ 21.17 ft
30 Degrees
28 ft.
60 degrees A. J. Calculations:
long leg= √3 × 16 ft
long leg=16 √3
long leg≈27.71 ft
27.71ft +63 inches
27.71 ft +5.25
≈ 32.96 ft Tangent x =
opposite/adjacent
Tan 60= x ft/16 ft
16 (Tan 60) = x
x ≈ 27.71 ft
27.71 ft + 5.25 ft
≈ 32.96 ft
60 degrees
16 ft
45 degrees• Valerie
• Special Right Triangles
• Calculation:
hypotenuse = √2 × leg
hypotenuse= 24√2
≈ 33.94 feet
33.94 feet + 5 feet
≈ 38.94 ft
• Tangent x= opposite / adjacent
tan 45 = x / 24
24( tan 45) = x
x ≈ 38.94 ft
45 degrees
24 ft
50 degrees Adam Calculations: Long leg = √3 × short leg
18 = √3 × a
18 / √3 = a
≈ 10.39 + 5.25
≈ 15.64 ft Tangent x = opposite / adjacent
tan 50 = a / 18
18 (tan 50) = a
a ≈ 15.64 ft18 ft
50 degrees
• We each measured a different angle of the building and counted the distance between us and the structure.
• We then used formulas for special right triangles and trigonometry to determine approximately how tall the building is.
•By adding the four averages :
32.96 ft + 38.94 ft + 21.17 ft + 15.64=108.71
then divide by four:
108.71 / 4 = 27.18 ft
• In conclusion, the building is approximately 27.18 ft tall.
Conclusion Slide