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**HallsofIvy** You have an isosceles triangle with sides of length 40, 40, and 64. Dropping a perpendicular divides that into two right triangles with hypotenuse of length 20 and horizontal leg of length 32 so the vertical leg has length $\displaystyle \sqrt{32^2- 20^2}= \sqrt{624}= 4\sqrt{39}$. Taking the tension in each string to be 'T', the vertical and horizontal components are $\displaystyle \frac[4\sqrt{39}}{40}T= \frac{\sqrt{39}}{10}T$ and $\displaystyle \frac{32}{40}T= \frac{4}{5}T$. The total of the two vertical components is $\displaystyle \frac{\sqrt{39}}{5}T$ and that must be equal to the weight, w. That is, $\displaystyle T= \frac{5}{\sqrt{39}}w $, **not** 5/6 w, though it is pretty close, 0.801w as compared to 5/6w= 0.833w.