MATLAB 已知3点求夹角>> x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;theta=acosd(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])))theta =47.8696>> x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;theta=acos(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,
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MATLAB 已知3点求夹角>> x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;theta=acosd(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])))theta =47.8696>> x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;theta=acos(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,
MATLAB 已知3点求夹角
>> x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;
theta=acosd(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])))
theta =
47.8696
>>
x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;
theta=acos(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])))
theta =
0.8355
>> pi/4
ans =
0.7854
为什么上面的3个点明显夹角是45度,但是算出来确实 47.8696呢
问题已解决来人送分
MATLAB 已知3点求夹角>> x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;theta=acosd(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,y1-y2])*norm([x1-x2,y3-y2])))theta =47.8696>> x1=1;y1=1;x2=0;y2=0;x3=0;y3=3;theta=acos(dot([x1-x2,y1-y2],[x3-x2,y3-y2])/(norm([x1-x2,
你给出的3个点的夹角就是47.8度,你应该该y3=2
。。。
谢谢楼主
我平时做法:
%A,B,C为三角形三定点,a,b,c为三边
A=[1,1];
B=[0,0];
C=[0,3];
a=sqrt((C(1)-B(1))^2+(C(2)-B(2))^2);
b=sqrt((A(1)-C(1))^2+(A(2)-C(2))^2);
c=sqrt((A(1)-B(1))^2+(A(2)-B(2)...
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谢谢楼主
我平时做法:
%A,B,C为三角形三定点,a,b,c为三边
A=[1,1];
B=[0,0];
C=[0,3];
a=sqrt((C(1)-B(1))^2+(C(2)-B(2))^2);
b=sqrt((A(1)-C(1))^2+(A(2)-C(2))^2);
c=sqrt((A(1)-B(1))^2+(A(2)-B(2))^2);
cosA=(c^2+b^2-a^2)/(2*b*c);
cosB=(a^2+c^2-b^2)/(2*a*c);
cosC=(a^2+b^2-c^2)/(2*a*b);
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