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