blocks|key|4148376|text|离题？但是你可以用余弦定律来做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4148377|找出A和B之间的距离(称为x)，B和C之间的距离(称为y)，以及A和C之间的距离(称为z)。|4148378|那么你就知道z%5E2=x%5E2%2By%5E2-2*x_y_cos(你想要的角度)|4148379|因此，角度是cos%5E-1((z%5E2-x%5E2-y%5E2)/(2xy))=角度|4148380|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|L|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|M|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|I|$]]

Off topic? But you can do it with the law of cosines:

Find the distance between A and B (call this x), and the distance between B and C (call this y), and the distance between A and C (call this z).

Then you know that z^2=x^2+y^2-2*xycos(ANGLE YOU WANT)

therefore, that angle is cos^-1((z^2-x^2-y^2)/(2xy))=ANGLE

blocks|key|888882|text|β=+arccos((a%5E2+%2B+c%5E2+-+b%5E2)+/+2ac)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|888883|其中a是侧对角角α，b是对角角β，c是对角角γ。所以β就是你们所说的角度ABC。|888884|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

β = arccos((a^2 + c^2 - b^2) / 2ac)

where a is the side opposite angle α, b is opposite angle β, and c is opposite angle γ. So β is what you called angle ABC.

blocks|key|2740208|text|float+angba+=+atan((a.y+-+b.y)+/+(a.x+-+b.x));
float+angbc+=+atan((c.y+-+b.y)+/+(c.x+-+b.y));
float+rslt+=+angba+-+angbc;
float+rs+=+(rslt+*+180)+/+3.141592;|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|2740209|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>float angba = atan((a.y - b.y) / (a.x - b.x));
float angbc = atan((c.y - b.y) / (c.x - b.y));
float rslt = angba - angbc;
float rs = (rslt * 180) / 3.141592;
</code></pre>

blocks|key|1635557|text|使用arccos的方法是危险的，因为我们可能会使它的参数等于，比如说，1.0000001，并以EDOMAIN错误结束。即使是atan方法也是危险的，因为它涉及除法，这可能导致除以零错误。最好使用atan2，将dx和dy值传递给它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1635558|entityMap^0|2|6|1B|7|1Q|4|2P|5|2W|2|2Z|2|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|G|$]]

Approach with <code>arccos</code> is dangerous, because we risk to have it's argument equal to, say, 1.0000001 and end up with <code>EDOMAIN</code> error. Even <code>atan</code> approach is dangerous, because it involves divisions, which may lead to division by zero error. Better use <code>atan2</code>, passing <code>dx</code> and <code>dy</code> values to it.

blocks|key|4148503|text|关于你的方法的第一个建议：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4148504|您所说的ac实际上是cb。但这没关系，这才是真正需要的。接下来，|offset|length|style|CODE|4148505|float+dotabac+=+(ab.x+*+ab.y+%2B+ac.x+*+ac.y);|code-block|syntax|javascript|4148506|这是你的第一个错误。两个向量的实数点积是：|BOLD|4148507|float+dotabac+=+(ab.x+*+ac.x+%2B+ab.y+*+ac.y);|4148508|现在,|4148509|float+rslt+=+acos(dacos);|4148510|这里，您应该注意到，由于计算过程中的一些精度损失，理论上dacos可能会变得大于1(或小于-1)。因此，您应该明确地检查这一点。|4148511|另外还有一个性能说明:为了计算两个向量的长度，您需要调用一个繁重的sqrt函数两次。然后将点积除以这些长度。相反，您可以对两个向量的长度平方的乘积调用sqrt。|4148512|最后，您应该注意到，您的结果在sign之前是准确的。也就是说，您的方法无法区分20°和-20°，因为两者的余弦是相同的。您的方法将为ABC和CBA产生相同的角度。|4148513|计算角度的一种正确方法是"oslvbo“：|4148514|float+angba+=+atan2(ab.y,+ab.x);
float+angbc+=+atan2(cb.y,+cb.x);
float+rslt+=+angba+-+angbc;
float+rs+=+(rslt+*+180)+/+3.141592;|4148515|(我刚刚用atan2替换了atan+)。|4148516|这是最简单的方法，它总是产生正确的结果。这种方法的缺点是，您实际上需要调用一个繁重的三角函数atan2两次。|4148517|我建议使用以下方法。它有点复杂(需要一些三角技巧才能理解)，但从性能的角度来看，它是优越的。它只调用一次三角函数atan2。而且没有平方根计算。|4148518|int+CGlEngineFunctions::GetAngleABC(+POINTFLOAT+a,+POINTFLOAT+b,+POINTFLOAT+c+)
{
++++POINTFLOAT+ab+=+{+b.x+-+a.x,+b.y+-+a.y+};
++++POINTFLOAT+cb+=+{+b.x+-+c.x,+b.y+-+c.y+};

++++//+dot+product++
++++float+dot+=+(ab.x+*+cb.x+%2B+ab.y+*+cb.y);

++++//+length+square+of+both+vectors
++++float+abSqr+=+ab.x+*+ab.x+%2B+ab.y+*+ab.y;
++++float+cbSqr+=+cb.x+*+cb.x+%2B+cb.y+*+cb.y;

++++//+square+of+cosine+of+the+needed+angle++++
++++float+cosSqr+=+dot+*+dot+/+abSqr+/+cbSqr;

++++//+this+is+a+known+trigonometric+equality:
++++//+cos(alpha+*+2)+=+[+cos(alpha)+]%5E2+*+2+-+1
++++float+cos2+=+2+*+cosSqr+-+1;

++++//+Here's+the+only+invocation+of+the+heavy+function.
++++//+It's+a+good+idea+to+check+explicitly+if+cos2+is+within+[-1+..+1]+range

++++const+float+pi+=+3.141592f;

++++float+alpha2+=
++++++++(cos2+<=+-1)+?+pi+:
++++++++(cos2+>=+1)+?+0+:
++++++++acosf(cos2);

++++float+rslt+=+alpha2+/+2;

++++float+rs+=+rslt+*+180.+/+pi;


++++//+Now+revolve+the+ambiguities.
++++//+1.+If+dot+product+of+two+vectors+is+negative+-+the+angle+is+definitely
++++//+above+90+degrees.+Still+we+have+no+information+regarding+the+sign+of+the+angle.

++++//+NOTE:+This+ambiguity+is+the+consequence+of+our+method:+calculating+the+cosine
++++//+of+the+double+angle.+This+allows+us+to+get+rid+of+calling+sqrt.

++++if+(dot+<+0)
++++++++rs+=+180+-+rs;

++++//+2.+Determine+the+sign.+For+this+we'll+use+the+Determinant+of+two+vectors.

++++float+det+=+(ab.x+*+cb.y+-+ab.y+*+cb.y);
++++if+(det+<+0)
++++++++rs+=+-rs;

++++return+(int)+floor(rs+%2B+0.5);


}|4148519|编辑：|4148520|最近，我一直在研究一个相关的主题。然后我意识到有一个更好的方法。这实际上是或多或少相同的(幕后)。然而，它更简单明了。|4148521|其思想是旋转这两个矢量，以便第一个矢量与(正)X方向对齐。显然，旋转这两个矢量不会影响它们之间的角度。在这样的旋转之后，我们只需要找出第二个矢量相对于X轴的角度。这正是atan2的作用所在。|4148522|旋转是通过将向量乘以以下矩阵来实现的：|4148523|4148524|a.x，a.y|unordered-list-item|4148525|-a.y，a.x|4148526|4148527|可以看到向量a乘以这样的矩阵确实朝着正X轴旋转。|4148528|注：严格地说，上面的矩阵不仅仅是旋转，它还在缩放。但在我们的例子中这是可以的，因为唯一重要的是向量的方向，而不是它的长度。|4148529|旋转后的向量b变为：|4148530|交叉b|4148531|-a.y+*+b.x+%2B+a.x+*+b.y+=+a+b|4148532|
|4148533|-a.y*+b.x+%2Ba.x*+b.y+=a|4148534|4148535|最后，答案可以表示为：|4148536|int+CGlEngineFunctions::GetAngleABC(+POINTFLOAT+a,+POINTFLOAT+b,+POINTFLOAT+c+)
{
++++POINTFLOAT+ab+=+{+b.x+-+a.x,+b.y+-+a.y+};
++++POINTFLOAT+cb+=+{+b.x+-+c.x,+b.y+-+c.y+};

++++float+dot+=+(ab.x+*+cb.x+%2B+ab.y+*+cb.y);+//+dot+product
++++float+cross+=+(ab.x+*+cb.y+-+ab.y+*+cb.x);+//+cross+product

++++float+alpha+=+atan2(cross,+dot);

++++return+(int)+floor(alpha+*+180.+/+pi+%2B+0.5);
}|4148537|entityMap^0|0|4|2|A|2|0|0|F|2|0|0|0|0|S|5|0|X|4|23|4|0|F|4|0|0|0|5|5|D|4|0|1A|5|0|1K|5|0|0|0|3|0|0|2C|5|0|0|0|0|0|0|6|1|0|0|2|0|6|1|0|2|1|0|P|1|R|1|0|0|L|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|2D|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|2E|8|@$D|2F|E|2G|F|G]|$D|2H|E|2I|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|2J|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|2K|8|@$D|2L|E|2M|F|O]]|9|@]|A|$]]|$1|P|3|Q|5|J|7|2N|8|@]|9|@]|A|$K|L]]|$1|R|3|S|5|6|7|2O|8|@]|9|@]|A|$]]|$1|T|3|U|5|J|7|2P|8|@]|9|@]|A|$K|L]]|$1|V|3|W|5|6|7|2Q|8|@$D|2R|E|2S|F|G]]|9|@]|A|$]]|$1|X|3|Y|5|6|7|2T|8|@$D|2U|E|2V|F|G]|$D|2W|E|2X|F|G]]|9|@]|A|$]]|$1|Z|3|10|5|6|7|2Y|8|@$D|2Z|E|30|F|G]]|9|@]|A|$]]|$1|11|3|12|5|6|7|31|8|@]|9|@]|A|$]]|$1|13|3|14|5|J|7|32|8|@]|9|@]|A|$K|L]]|$1|15|3|16|5|6|7|33|8|@$D|34|E|35|F|G]|$D|36|E|37|F|G]]|9|@]|A|$]]|$1|17|3|18|5|6|7|38|8|@$D|39|E|3A|F|G]]|9|@]|A|$]]|$1|19|3|1A|5|6|7|3B|8|@$D|3C|E|3D|F|G]]|9|@]|A|$]]|$1|1B|3|1C|5|J|7|3E|8|@]|9|@]|A|$K|L]]|$1|1D|3|1E|5|6|7|3F|8|@$D|3G|E|3H|F|O]]|9|@]|A|$]]|$1|1F|3|1G|5|6|7|3I|8|@]|9|@]|A|$]]|$1|1H|3|1I|5|6|7|3J|8|@$D|3K|E|3L|F|G]]|9|@]|A|$]]|$1|1J|3|1K|5|6|7|3M|8|@]|9|@]|A|$]]|$1|1L|3|-4|5|6|7|3N|8|@]|9|@]|A|$]]|$1|1M|3|1N|5|1O|7|3O|8|@]|9|@]|A|$]]|$1|1P|3|1Q|5|1O|7|3P|8|@]|9|@]|A|$]]|$1|1R|3|-4|5|6|7|3Q|8|@]|9|@]|A|$]]|$1|1S|3|1T|5|6|7|3R|8|@$D|3S|E|3T|F|G]]|9|@]|A|$]]|$1|1U|3|1V|5|6|7|3U|8|@$D|3V|E|3W|F|O]]|9|@]|A|$]]|$1|1W|3|1X|5|6|7|3X|8|@$D|3Y|E|3Z|F|G]]|9|@]|A|$]]|$1|1Y|3|1Z|5|6|7|40|8|@$D|41|E|42|F|O]]|9|@]|A|$]]|$1|20|3|21|5|1O|7|43|8|@$D|44|E|45|F|O]|$D|46|E|47|F|O]]|9|@]|A|$]]|$1|22|3|23|5|6|7|48|8|@]|9|@]|A|$]]|$1|24|3|25|5|1O|7|49|8|@$D|4A|E|4B|F|O]]|9|@]|A|$]]|$1|26|3|-4|5|6|7|4C|8|@]|9|@]|A|$]]|$1|27|3|28|5|6|7|4D|8|@]|9|@]|A|$]]|$1|29|3|2A|5|J|7|4E|8|@]|9|@]|A|$K|L]]|$1|2B|3|-4|5|6|7|4F|8|@]|9|@]|A|$]]]|2C|$]]

First suggestions regarding your method:

What you call <code>ac</code> is actually <code>cb</code>. But it's ok, this is what really needed.
Next,

<pre><code>float dotabac = (ab.x * ab.y + ac.x * ac.y);
</code></pre>

This is your first mistake. The real dot product of two vectors is:

<pre><code>float dotabac = (ab.x * ac.x + ab.y * ac.y);
</code></pre>

Now, 

<pre><code>float rslt = acos(dacos);
</code></pre>

Here you should note that due to some precision loss during the calculation it's theoretically possible that <code>dacos</code> will become bigger than 1 (or lesser than -1). Hence - you should check this explicitly.

Plus a performance note: you call a heavy <code>sqrt</code> function twice for calculating the length of two vectors. Then you divide the dot product by those lengths.
Instead you could call <code>sqrt</code> on the multiplication of squares of length of both vectors.

And lastly, you should note that your result is accurate up to the <code>sign</code>. That is, your method won't distinguish 20° and -20°, since the cosine of both are the same.
Your method will yield the same angle for ABC and CBA.

One correct method for calculating the angle is as "oslvbo" suggests:

<pre><code>float angba = atan2(ab.y, ab.x);
float angbc = atan2(cb.y, cb.x);
float rslt = angba - angbc;
float rs = (rslt * 180) / 3.141592;
</code></pre>

(I've just replaced <code>atan</code> by <code>atan2</code>).

It's the simplest method, which always yields the correct result. The drawback of this method is that you actually call a heavy trigonometry function <code>atan2</code> twice.

I suggest the following method. It's a bit more complex (requires some trigonometry skills to understand), however it's superior from the performance point of view.
It just calls once a trigonometry function <code>atan2</code>. And no square root calculations.

<pre><code>int CGlEngineFunctions::GetAngleABC( POINTFLOAT a, POINTFLOAT b, POINTFLOAT c )
{
 POINTFLOAT ab = { b.x - a.x, b.y - a.y };
 POINTFLOAT cb = { b.x - c.x, b.y - c.y };

 // dot product 
 float dot = (ab.x * cb.x + ab.y * cb.y);

 // length square of both vectors
 float abSqr = ab.x * ab.x + ab.y * ab.y;
 float cbSqr = cb.x * cb.x + cb.y * cb.y;

 // square of cosine of the needed angle 
 float cosSqr = dot * dot / abSqr / cbSqr;

 // this is a known trigonometric equality:
 // cos(alpha * 2) = [ cos(alpha) ]^2 * 2 - 1
 float cos2 = 2 * cosSqr - 1;

 // Here's the only invocation of the heavy function.
 // It's a good idea to check explicitly if cos2 is within [-1 .. 1] range

 const float pi = 3.141592f;

 float alpha2 =
 (cos2 &lt;= -1) ? pi :
 (cos2 &gt;= 1) ? 0 :
 acosf(cos2);

 float rslt = alpha2 / 2;

 float rs = rslt * 180. / pi;


 // Now revolve the ambiguities.
 // 1. If dot product of two vectors is negative - the angle is definitely
 // above 90 degrees. Still we have no information regarding the sign of the angle.

 // NOTE: This ambiguity is the consequence of our method: calculating the cosine
 // of the double angle. This allows us to get rid of calling sqrt.

 if (dot &lt; 0)
 rs = 180 - rs;

 // 2. Determine the sign. For this we'll use the Determinant of two vectors.

 float det = (ab.x * cb.y - ab.y * cb.y);
 if (det &lt; 0)
 rs = -rs;

 return (int) floor(rs + 0.5);


}
</code></pre>

EDIT:

Recently I've been working on a related subject. And then I've realized there's a better way. It's actually more-or-less the same (behind the scenes). However it's more straightforward IMHO.

The idea is to rotate both vectors so that the first one is aligned to (positive) X-direction. Obviously rotating both vectors doesn't affect the angle between them. OTOH after such a rotation one just has to find out the angle of the 2nd vector relative to X-axis. And this is exactly what <code>atan2</code> is for.

Rotation is achieved by multiplying a vector by the following matrix:

<ul>
<li>a.x, a.y</li>
<li>-a.y, a.x</li>
</ul>

Once may see that vector <code>a</code> multiplied by such a matrix indeed rotates toward positive X-axis.

Note: Strictly speaking the above matrix isn't just rotating, it's also scaling. But this is ok in our case, since the only thing that matters is the vector direction, not its length.

Rotated vector <code>b</code> becomes:

<ul>
<li>a.x * b.x + a.y * b.y = a dot b</li>
<li>-a.y * b.x + a.x * b.y = a cross b</li>
</ul>

Finally, the answer can be expressed as:

<pre><code>int CGlEngineFunctions::GetAngleABC( POINTFLOAT a, POINTFLOAT b, POINTFLOAT c )
{
 POINTFLOAT ab = { b.x - a.x, b.y - a.y };
 POINTFLOAT cb = { b.x - c.x, b.y - c.y };

 float dot = (ab.x * cb.x + ab.y * cb.y); // dot product
 float cross = (ab.x * cb.y - ab.y * cb.x); // cross product

 float alpha = atan2(cross, dot);

 return (int) floor(alpha * 180. / pi + 0.5);
}
</code></pre>

blocks|key|2740302|text|以下是计算直角值的一种快速而正确的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2740303|double+AngleBetweenThreePoints(POINTFLOAT+pointA,+POINTFLOAT+pointB,+POINTFLOAT+pointC)
{
++++float+a+=+pointB.x+-+pointA.x;
++++float+b+=+pointB.y+-+pointA.y;
++++float+c+=+pointB.x+-+pointC.x;
++++float+d+=+pointB.y+-+pointC.y;

++++float+atanA+=+atan2(a,+b);
++++float+atanB+=+atan2(c,+d);

++++return+atanB+-+atanA;
}+|code-block|syntax|javascript|2740304|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is a quick and correct way of calculating the right angle value:

<pre><code>double AngleBetweenThreePoints(POINTFLOAT pointA, POINTFLOAT pointB, POINTFLOAT pointC)
{
 float a = pointB.x - pointA.x;
 float b = pointB.y - pointA.y;
 float c = pointB.x - pointC.x;
 float d = pointB.y - pointC.y;

 float atanA = atan2(a, b);
 float atanB = atan2(c, d);

 return atanB - atanA;
} 
</code></pre>

blocks|key|4148581|text|下面是获取3个点(A，B，C)之间的角度的OpenCV方法，其中B是顶点：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4148582|int+getAngleABC(+cv::Point2d+a,+cv::Point2d+b,+cv::Point2d+c+)
{
++++cv::Point2d+ab+=+{+b.x+-+a.x,+b.y+-+a.y+};
++++cv::Point2d+cb+=+{+b.x+-+c.x,+b.y+-+c.y+};

++++float+dot+=+(ab.x+*+cb.x+%2B+ab.y+*+cb.y);+//+dot+product
++++float+cross+=+(ab.x+*+cb.y+-+ab.y+*+cb.x);+//+cross+product

++++float+alpha+=+atan2(cross,+dot);

++++return+(int)+floor(alpha+*+180.+/+M_PI+%2B+0.5);
}|code-block|syntax|javascript|4148583|基于@valdo的优秀解决方案|4148584|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Here's an OpenCV way to get the angle between 3 points (A, B, C) with B as the vertex:

<pre><code>int getAngleABC( cv::Point2d a, cv::Point2d b, cv::Point2d c )
{
 cv::Point2d ab = { b.x - a.x, b.y - a.y };
 cv::Point2d cb = { b.x - c.x, b.y - c.y };

 float dot = (ab.x * cb.x + ab.y * cb.y); // dot product
 float cross = (ab.x * cb.y - ab.y * cb.x); // cross product

 float alpha = atan2(cross, dot);

 return (int) floor(alpha * 180. / M_PI + 0.5);
}
</code></pre>

Based on the excellent solution by @valdo

Given points ABC, how could I find angle ABC? I'm making a feehand tool for a vector drawing application and to minimize the number of points it generates, I wont add a points unless the angle of the mouse position and the last 2 points is greater than a certain threshold.
Thanks

what I had:

<pre><code>int CGlEngineFunctions::GetAngleABC( POINTFLOAT a, POINTFLOAT b, POINTFLOAT c )
{
 POINTFLOAT ab;
 POINTFLOAT ac;

 ab.x = b.x - a.x;
 ab.y = b.y - a.y;

 ac.x = b.x - c.x;
 ac.y = b.y - c.y;

 float dotabac = (ab.x * ab.y + ac.x * ac.y);
 float lenab = sqrt(ab.x * ab.x + ab.y * ab.y);
 float lenac = sqrt(ac.x * ac.x + ac.y * ac.y);

 float dacos = dotabac / lenab / lenac;

 float rslt = acos(dacos);
 float rs = (rslt * 180) / 3.141592;
 RoundNumber(rs);
 return (int)rs;


}
</code></pre>

Angle between 3 points?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

给定点ABC，我如何找到角度ABC？我正在为一个矢量绘图应用程序制作一个feehand工具，为了最小化它生成的点的数量，我不会添加一个点，除非鼠标位置和最后两个点的角度大于某个阈值。谢谢我所拥有的：int CGlEngineFunctions::GetAngleABC( POINTFLOAT a, POINTFLOAT...

问三个点之间的角度？
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问三个点之间的角度？EN