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A similar question has been asked before, but because of float imprecision when calculating $\arccos$ and comparing with $\pi,$ I would like to know given the input with points $A, B, C$ (in a given order), without using any trigonometric functions, how (if it is possible) can you find if an angle is acute or obtuse (or right)?

Community
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Jack Pan
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2 Answers2

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The angle will be acute if and only if the dot product is positive.

John Wayland Bales
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  • Is there a proof I can find somewhere? This is exactly what I was looking for btw, thanks. – Jack Pan Jun 30 '16 at 21:59
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    If you place an angle $\theta$ with its vertex at the center of the unit circle with its initial side along the positive $x$-axis, then the terminal side of the angle crosses the unit circle at the point with coordinates $(\cos\theta,\sin\theta).$ So if the angle is acute, then the $x$-coordinate of that point--that is to say-- $\cos\theta$ will be positive because the point will be in the first quadrant. If the terminal side is in quadrant II, then $x$ will be negative. In the cosine formula, only the dot-product can change sign because the denominator in the formula is always non-negative. – John Wayland Bales Jun 30 '16 at 22:15
  • Thank you! One last thing, can you show me out to work out this example with $∠BCA$ with $B=(0, 2), C=(2, 0), A=(0, 0),$ I get the dot product is equal to $(2-0)(0-2)+(0-2)(0-0)=-4.$ But it seems to be acute. – Jack Pan Jun 30 '16 at 22:17
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    The initial point of the vectors must be the vertex! So the two vectors are $\vec{CB}$ and $\vec{CA}$. And $\vec{CB}=(-2,2)$ and $\vec{CA}=(-2,0)$. – John Wayland Bales Jun 30 '16 at 22:26
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The sign of $\vec{AB}\cdot \vec{AC}$ tell you whether the angle at $A$ is acute (+ve), right ($0$) or obtuse (-ve).

Joffan
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  • Thank you! One last thing, can you show me out to work out this example with $∠BCA$ with $B=(0, 2), C=(2, 0), A=(0, 0),$ I get the dot product is equal to $(2-0)(0-2)+(0-2)(0-0)=-4.$ But it seems to be acute. – Jack Pan Jun 30 '16 at 22:16
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    $\vec{CB} = (-2,2)$ and $\vec{CA} = (-2,0)$ . So $\vec{CB}\cdot \vec{CA} = 4+0 = 4 > 0 \implies$ acute. – Joffan Jun 30 '16 at 22:34