A seven foot ladder leaning against wall touches wall x feet above the ground - write eqtn. in terms of x...?

Question

A seven foot ladder leaning against wall touches wall x feet above the ground - write eqtn. in terms of x...for the distance from the foot of the ladder to the base of the wall.

Erm...how is this done? I've been working at this for about 45 minutes. I swear law of cosines is involved. Here's what I have (let the unknown distance of ladder base to wall equal y)

Since the law of cosines states:

a^2 = b^2 + c^2 - 2bc cos(A)...

I derived...

y^2 = x^2 + 7^2 - 2(7)(x) cos(x/7)

Therefore...

y = sqrt[x^2 + 49 - 14x cos(x/7)]

For some reasons the answer don't check out with test data that uses the Pythagorean theorem? What gives?

Answer 1

Zachman, I'd say that 7x = x + 2.14 (pi) times the unknown's right triangle's circumference = the distance of the radii's (radius plural) diameter.  Good luck.

Answer 2

Er, I ended up with y= 7-x, but that doesn't make much sense.

Hmp. It's been a long summer apparently.

This was my work anyway.

y^2= x^2 + 7^2 - (2)(x)(7)(x/7)

y^2= x^2 + 49 - (14x)(x/7)

y^2= x^2 + 49 - (2x^2)

Then square root of both sides, so

y= sqrt[y^2= 49 -x^2]

y=7-x.

Wow. Good luck. I don't remember.

Answer 3

alright i believe your WAY overthinking it! first draw a right triangle. the hypotenuse, or ladder, is 7 feet. the vertical side of the triangle (the wall) is x. the horizontal side (distance from the foot of the ladder to the base of the wall) is y.

by the pythagorean theorem, y^2 + x^2 = 49.

then just solve for y.

y^2 = 49 - x^2.

y= sqrt(49-x^2).

and you're done! you have an equation for the distance, y, in terms of x.

Answer 4

im sorry, but i hear yah brotha!!!  

Answer 5

1) Get jackhammer.

2) Break down wall.

3) Equation is unsolvable. No wall for ladder to touch.

4) Problem solved.