I´d say "d" is the distance from the eye to the wall.
Now substracting 1.2-1 you´ll get the distance of the wall of the smallest triangle = 0.2 And you do 1.5-0.2= 0.3 that´s the distance of the wall of the other triangle. Then you solve everything with Pitagoras theorem. You have 2 rectangle triangles.
B+alfa=45°
tan^-1(0.2/d)=B
tan^-1(1.3/d)=alfa
THEN:
tan^-1(0.2/d)+tan^-1(1.3/d)=45°
Now you have 3 ecs and 3 variables.
alfa,B and "d"
I believe it is positive so option 2
Hello,
Please, see the attached file.
Thanks.
We have to find the GCD between 10, 16 and 4 and between x^5, x^4 and x^2
GCD (10,16,4) = 2
GCD (x^5,x^4,x^2) = x^2
So we divide all terms for 2x^2
Final result: 2x^2(5x^3-8x^2+2)
