Calculating Vector Cosines: Understanding the Proper Method

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I don't understand how they got these values for the cosines of the vector.
My understanding is that the angle between the unit vector and the x,y and z axises are used to determine the cosine of the vector, so for example, the vector creates a 45 degree angle with the x-axis and so the vector cosine for that axis is cos(180-45), but this doesn't give me the correct answer, so can someone tell me where I'm going wrong?
(I got 45degrees because the dome on top of the hut is a square and half of a right angle is 45 degrees)
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Ry122 said:
so can someone tell me where I'm going wrong?
You're taking the ratio of the projection in the x-axis direction to the projection in the x,z plane.
Instead you should just to take the ratio of the projection in the x-axis direction to the total vector length (and similarly for the other two direction cosines).

For example the projection in the x-axis direction is 10m. The projection in the x,z plane is 10 \, \sqrt{2}. You seem to be using 10 / (10 \, \sqrt{2}) when you should be using 10/16.2.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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