Bekah Sean Griffen Kohta 5 th period 60 Degrease Bekah Tan x= Opposite/adjacent Tan 60= x feet / 14 feet 14(tan 60)= X 24.249 feet+ 14 feet 5.125 feet+

How Tall Is IT? BekahSean

Griffen Kohta

5th period

60 DegreaseBekah Tan x=

Opposite/adjacent

Tan 60= x feet / 14 feet14(tan 60)= X24.249 feet+ 14 feet5.125 feet+ 24.249 feet =

29.374 feet

Long leg= square root of 3 x short leg

X = square root of 3 x 14 feet

X= 14 x square root of 3/ 3

X= 24.25 feet60 degrees

14 feet 5 feet

45 Degrees Kohta 60 inches (eye height) 28 feet base

Tan x = opposite adjacentTan (45)= x feet 28 feet28 (tan 45)= x

28 feet = x28 feet + 60 inches

28 feet + 5 feet = 33 feet

45-45-90

Side=Side

28=28

Side+Height=Pole Height

28+5=33feet

5 Feet

28 feet

30 Degrees Sean

Tan x = opposite / adjacent

Tan30 = x / 38ft

38(Tan30) = x

X ≈21.94 + 5.75

X≈27.69 ft

38 ft5.75

Long leg =√3 ×short leg

38 = √3 × s. leg

s. Leg = 38√3 / 3

s. Leg ≈ 21.94 + 5.75

X ≈ 27.69 ft.

30 degrees

X

24 Degrees Griffin

24 Degrees

67 Feet

5.75 Feet

Tan x ≈ opposite Adjacent

Tan (24) ≈ X Feet 67 Feet

67 [Tan (24)] ≈ X Feet

X ≈ 29.83 Feet

29.83 Feet + 5.75 Feet ≈ 35.58 Feet

So………What Did We Learn Again?We noted that as the measure of the angle increased, the

closer we got to the object. This made the angle sharper, and obviously decreased the distance from us to the light pole.

The height of the light pole stayed the same We also noted that our light pole was crooked, so the

measurements were most likely inaccurate. Also, one can not get an accurate measurement by simply taking steps.