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.
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.
Let x = CX. .Then AX = 8 - x.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