In right triangle ABC with right angle ACB,AC=8 inches and BC=6 inches point x in on segment AC equidistant from A and B . Find CX.

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- Mar 21st 2007, 03:50 PMGodfatherRight Triangles
In right triangle ABC with right angle ACB,AC=8 inches and BC=6 inches point x in on segment AC equidistant from A and B . Find CX.

- Mar 21st 2007, 04:40 PMtopsquark
I'll give you the method.

Note that triangle AXB is isosceles. Thus angle ABX is equal to angle BAX. We can find angle BAX since triangle ABC is right, using the inverse tangent function. We also can find angle ABC because of the same fact. Thus by subtraction we can find angle XBC. We now know the value of the angle opposite the side XC of the right triangle XBC. And we know the length of the side adjacent to the angle. So we can use the tangent function to find the length of XC.

-Dan - Mar 28th 2007, 03:43 PMRimas
- Mar 28th 2007, 04:11 PMajacka
easy.. basically, u know that AX and BX are equal, so they r both 8. so then you know 2 sides of the triangle BXC, so all you're left with is do a simple pythagorean theorem. CX^2 = BX^2 - BC^2.

- Mar 28th 2007, 04:58 PMtopsquark
- Mar 28th 2007, 05:09 PMajacka
oops i misread the question, the diagram show AX = 8 hmm so..

- Mar 29th 2007, 05:21 PMSoroban
Hello, Godfather!

Quote:

In right triangle ABC with right angle ACB,AC=8 inches and BC=6 inches,

point X in on segment AC equidistant from A and B. .Find CX.

In right triangle XCB, we have: .x² + 6² .= .BX²

Since BX = AX, we have: .x² + 36 .= .(8 - x)²

Then: .x² + 36 .= .64 - 16x + x² . → . 16x = 28 . → . x = 7/4