- #1
Matejxx1
- 72
- 1
Homework Statement
ok so my professor gave me this problem to solve, it goes like this :(I will also have a picture below)
In the square (ABCD) is a point P which divides the side BC into 2 halves and point R which divides the side CD into 2 halves
The angles at APB and ARB are the same.
Calculate the angle x and the ratio of sides of the square
2. Homework Equations
he gave me an answer x=arc tan (2*sqrt(2))
and
a:b=sqrt(2):1
I got those answers but I also got a second one and I an wondering if you guys can tell me why the 2nd answer is not possible
3. The Attempt at a Solution
ok so first I started out by writting sin(x)
sin(x)= a/sqrt(a2+(b/2)2) i got that from the ABP triangle
sin(x)=a/sqrt(a2+b2/4)
sin(x)=a/sqrt((4a2+b2)/4)
sin(x)=2a/sqrt(4a2+b2)
secondly I calculated the length of the side AR
AR=sqrt(b2+a2/4)
AR=sqrt(4b2+a2)/2
thirdly I use the formula to calculate the area of the ARB triangle
Area=AR2*sin(x)/2
Area=(4b2+a2)*2a/(8*sqrt(4a2+b2))
Area=(4b2a+a3)/(4*sqrt(4a2+b2)
ok from now on I figured out that this are is equal to half of the area of a square
a*b/2=(4b2a+a3)/(4*sqrt(4a2+b2)
2*b*sqrt(4a2+b2)=4b2+a2
here i squared both sides
4b2*(4a2+b2)=16b4+8b2a2+a4
16a2b2+4b4=16b4+8b2a2+a4
0=12b4-8a2b2+a4
here i created a new variable m=b2
0=12m2-8m*a2+a4
m1,2=(8a2+(or)-sqrt(64a4-48a4))724
m1,2=(8a2+(or)-4[aSUP]2[/SUP])/24
m1=a2/2
m2=a2/6
ok so
b2=a2/2
and
b2=a2/6
I won't check for a2/2 because I already know that is the right answer. I'm just wondering why this is not the right answer as well
b2=a2/6
so know I finally returned to my initial equation
sin(x)=2a/sqrt(4a2+b2) and I put in b2
sin(x)=2a/sqrt(4a2+a2/6)
sin(x)=2a/sqrt(25a2/6)
sin(x)=2a*sqrt(6)/5a
sin(x)=2*sqrt(6)/5
and if I draw this triangle i can figure out the 3 side
and then I can calculate the angles as
tan(x)=(2*sqrt(6))/...x=arc tan (2*sqrt(6))
and
a:b=sqrt(6):1
can somebody help me find my mistake or tell me why this is not a possible answer ?
thanks
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