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$a \ne b$

  • Are the angles a and b as shown in the image equal?
  • Why (not)?
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They can't be equal unless the height is zero. One reason: each angle is the arc cosine of the dot product of unit vectors pointing along the two rays from the angle. – bgins May 3 '12 at 14:00
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Hard to tell. You gave no lengths or congruences to start with, and you never even said if the figure is a two-dimensional or a (projection of a) three-dimensional object. – J. M. May 3 '12 at 14:03
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If $2$D, no; if $3$D, maybe. – André Nicolas May 3 '12 at 14:19
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If this is a two-dimensional picture, then no, because vertex of $b$ is interior to the circumcircle of the external triangle. – lhf May 3 '12 at 14:21
If this triangle is inscribed in a circle, and the inside point is the center of the circle, then angle b is twice angle a. – GEdgar May 3 '12 at 14:24

closed as not a real question by Rahul Narain, t.b., Chris Eagle, Zev Chonoles♦ Jun 9 '12 at 20:15

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

1 Answer

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Imagine your solid lives between the origin $O=(0,0,0)$, $X=(x,0,0)$, $Y=(0,y,0)$ and $Z=(0,0,z)$ for some $x,y,z > 0$, with angles $a=\angle XZY$ and $b=\angle XOY=90^\circ=\frac{\pi}2$ radians. (I am assuming that the three lines which meet where angle $b$ is all meet at right angles.) There is a formula that says: $$ \eqalign{ \cos a &=\frac{ZX}{|ZX|}\cdot\frac{ZY}{|ZY|}\\ &=\frac{(x,0,-z)}{\sqrt{x^2+z^2}}\cdot\frac{(0,y,-z)}{\sqrt{y^2+z^2}}\\ &=\frac{z^2}{\sqrt{\left(x^2+z^2\right)\left(y^2+z^2\right)}}\\ &=\left[\left(1+\frac{x^2}{z^2}\right)\left(1+\frac{y^2}{z^2}\right)\right]^{-1/2}\\ } $$ which is clearly not equal to $\cos b=0$ for $x,y,z > 0$. In fact, $a < b$ for the same reason that objects on the ground appear smaller from an elevation.

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