********Can Someone PLEASE help me with my PreCalculus Homework??*******?

Question

John's paint crew knows from experience that its 18 ft ladder is parpticularly stable when the distance from the ground to the top of the ladder ir 5 ft more than the distance from the building to the base of the ladder as shown in the figure. (Sorry, can't get pic on here ) In this position, how far up the building does the ladder reach?

Answer 1

Here's the setup:

sqrt(x^2 + (x+5)^) = 18

If you need further help, add additional details.

Answer 2

Okay, when it's leaning against a building, the ladder forms a right triangle with a hypotenuse of known length (18 feet). The vertical leg of your triangle is 5 feet longer than the base. So let's just call the base (distance from the wall) X; therefore, the height of your triangle is X+5. Pythagoras tells us that any right triangle follows the rule a^2 + b^2 = c^2, so you can substitue numbers and variables and get:

X^2 + (X+5)^2 = 18^2 or
2X^2 +10X +25 = 324 or
2X^2 + 10X - 299 = 0 simplfying, you get
(2X + 23)(X - 13) = 0
X = 13 or -11.5

Obviously, the height can't be negative, so the ladder reaches 13 feet up the building.

Answer 3

The ladder, ground and building form a right trangle.

(side1)^2 + (side2)^2 = (hypotenuse)^2

(ground)^2 + (building)^2 = (ladder)^2

(x)^2 + (x + 5) = 18^2

x^2 + x^2 + 10x + 25 = 324
2x^2 + 10x - 299 = 0

x =[ -10 +/- sqrt(100 +2392) ] /4

x = 10 ft
So, distance up from ground = 15 feet

Answer 4

Use the pythagorean thereom. a^2 + b^2 = c^2.
In this case x^2 + (x+5)^2 = 18^2...where x is the distance from the building to the base of the ladder. Solve for x using whatever method you prefer.

Answer 5

Your initial math is correct. On inspection, Tan 2pi = tan 360 = 0. Cot 3 pi/2 = cot 270 = "-" 0 0 - 0 = 0. This also satisfies the equation.

Answer 6

18(squared) = H (squared) + [(H-5) (squared)]

Solve for H

USING THE QUADRATIC EQUATION METHOD:

H = (b^2 +- SQRT(b^2 - 4ac))/2a

THE THREE SIDES OF THE RIGHT TRIANGLE ARE (SOLVING BY THE QUADRATIC EQUATION METHOD)

VERTICAL: 15 FT
HORIZONTAL: 10 FT
ALONG THE LADDER: 18 FT

Answer 7

The answer by "theyuks" above is correct till the last step. The "factorization" step is wrong. The solution comes to 7.05

Answer 8

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