smilyfacebkwrm
New member
- Joined
- Oct 9, 2004
- Messages
- 7
Two buildings are d units apart. A ladder 20 units long has its foot resting against building 1 and its top against the side of building 2. A second ladder 15 units long has its foot against building 2 and its top against the side of building 1. The ladders touch each other at a point c units above the ground.
a) show that (1/root 4-d^2) + (1/root 225-d^2) = 1/c
b) if d = 12, use part (a) to find c
c) if c=8, find d. since it is difficult to solve for d directly, you can use your calculator to approxmiate d to the nearest hundreth by finding the value of d for which the expressiong (1/root 4-d^2) + (1/root 225-d^2) - 1/8 changes sign
I can't get a and so I don't understand b or c either. I think with a I'd be able to do b, but not c... I can't get anywhere and I have no clue where to start with this.
a) show that (1/root 4-d^2) + (1/root 225-d^2) = 1/c
b) if d = 12, use part (a) to find c
c) if c=8, find d. since it is difficult to solve for d directly, you can use your calculator to approxmiate d to the nearest hundreth by finding the value of d for which the expressiong (1/root 4-d^2) + (1/root 225-d^2) - 1/8 changes sign
I can't get a and so I don't understand b or c either. I think with a I'd be able to do b, but not c... I can't get anywhere and I have no clue where to start with this.