In square ABCD, we select a point P such that AP=a, BP=2a, CP=3a. Find ∠APB?
Let
A=(0,0)
B=(1,0)
C=(1,1)
D=(0,1)
P=(x,y)
a=sqrt(x^2+y^2)
sqrt((x-1)^2 + y^2) = 2*sqrt(x^2+y^2)
sqrt((x-1)^2 + (y-1)^2) = 3*sqrt(x^2+y^2)
(x-1)^2 + y^2 = 4(x^2+y^2)
(x-1)^2 + (y-1)^2 = 9(x^2+y^2)
(x-1)^2 + y^2 = 4(x^2+y^2),
(x-1)^2 + (y-1)^2 = 9(x^2+y^2)
Using Wolfram Alpha to solve this:
x = (1/17)*(1 + 3*sqrt(2))
y = (1/17)*(5*sqrt(2) - 4)
Program using different meanings of a and b:
x = (1/17)*(1 + 3*sqrt(2)) % of point P
y = (1/17)*(5*sqrt(2) - 4)
b=norm([x,y]); % opposite point B in small triangle APB
a=2*b; % opposite point A in small triangle APB
c=1; % length of AB
% c^2 = a^2 + b^2 - 2*a*b*cos(APB)
% cos(APB) = (a^2 + b^2 - c^2) / (2*a*b)
APB=acosd((a^2 + b^2 - c^2) / (2*a*b))
finds
x =
0.308390628654076
y =
0.180651047756793
APB =
135
Angle APB = 135°
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Posted by Charlie
on 2024-06-20 12:28:29 |
solution
