MATH SOLVE

4 months ago

Q:
# The height of an object from a distance

Accepted Solution

A:

y = height

x = distance from tower

tan(19.9) = y/x

tan(21.8) = y/(x-50) ( x is 50 feet closer so subtract 50 from x)

find the tan of the angles:

tan19.9 = 0.362

tan21.8 = 0.400

replace those values in the equations:

0.362 = y /x

0.400 = y / (x-50)

rewrite to solve for height:

0.362x = y

0.400(x-50) = y distribute the 0.400 and rewrite as 0.400x-20 =y

now solve:

0.362x = 0.400x-20

add 20 to each side:

0.362x +20 = 0.400x

subtract 0.362x from each side:

20 = 0.038x

divide both sides by 0.038

x = 20 / 0.038 = 526.3 feet

now that x is known replace x in the original equation and solve for y:

tan19.9 = y /x

0.362 = y / 526.3

y = 0.362 x 526.3

y = 190.5

the height of the tower is 190.5 feet

x = distance from tower

tan(19.9) = y/x

tan(21.8) = y/(x-50) ( x is 50 feet closer so subtract 50 from x)

find the tan of the angles:

tan19.9 = 0.362

tan21.8 = 0.400

replace those values in the equations:

0.362 = y /x

0.400 = y / (x-50)

rewrite to solve for height:

0.362x = y

0.400(x-50) = y distribute the 0.400 and rewrite as 0.400x-20 =y

now solve:

0.362x = 0.400x-20

add 20 to each side:

0.362x +20 = 0.400x

subtract 0.362x from each side:

20 = 0.038x

divide both sides by 0.038

x = 20 / 0.038 = 526.3 feet

now that x is known replace x in the original equation and solve for y:

tan19.9 = y /x

0.362 = y / 526.3

y = 0.362 x 526.3

y = 190.5

the height of the tower is 190.5 feet