Trigonometric equations (finding angles)

[Matejxx1]

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
m_1,2=(8a²+(or)-sqrt(64a⁴-48a⁴))724
m_1,2=(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

 

[Simon Bridge]

I can't tell your mistake bc I'm having trouble following your reasoning.
You check answers by drawing out the resulting rectangle and checking the angles etc make sense. This approach also helps you understand what happened.

I'd have divided side AB in two, labelling the halfway point Q
Then I'd use the tangent definition on the triangles ABP and RQB
... the tangent half-angle formula looks useful here.

 

[ehild]

Your approach allows the angle ARB greater than 90°, as you work entirely with sines. But the angle APB has to be less than 90°. The tangent of half of the angle ARB is a/(2b) and a/(2b)<1 must hold, that is a/b<2. Working with the half-angle formula as Simon suggested you get only a=√2 b. If you assume a=√6 b, you get negative value for the tangent of angle ARB, although its sine is the same as that of APB.
Draw also the triangle ABR.

 

[Matejxx1]

Thank you
just to check if I really get it
if I would allow that the tangent half angle at ARB is >1 than that would mean that the whole angle would be over 90° and since the picture only shows the posibility of the angle being less 90° the second solution is ruled out.?
Another question if I may.
What is the best way to check your solutions in problems like this ? is it to draw it out or do somethng else?

 

[ehild]

Thank you
just to check if I really get it
if I would allow that the tangent half angle at ARB is >1 than that would mean that the whole angle would be over 90° and since the picture only shows the posibility of the angle being less 90° the second solution is ruled out.?

Yes, correct.

Another question if I may.
What is the best way to check your solutions in problems like this ? is it to draw it out or do something else?

It is always necessary to investigate the possible range of the solution(s). Here it is clear that the angle should be less than 90°,so the half-angle is less than 45°. Also useful to draw it, but then it is not so evident that a solution is wrong.

 

[Simon Bridge]

I get ##a/b = \sqrt {6}## via half-angle formula. <sniff> :)

 

[ehild]

I get ##a/b = \sqrt {6}## via half-angle formula. <sniff> :)

How?
tan(x)=2a/b
tan(x/2)=a/(2b)
##\tan(x)=\frac{2\tan(x/2)}{1-\tan^2(x/2)}##, well, this is rather the double-angle formula, but it gives a/b=√2.