Pretest Questions
i'm basically done with the pretest except for these 4 questions. can someone please explain them to me?
10 and 14. i don't even know where to start. any suggestions?
11. i started by finding the diagonal on the bottom face: sqrt(2x^2), if the sides of the cube=x and labeling the angle ABC theta.
tan(theta)=x/sqrt(2x^2)
--> theta=tan^-1(x/sqrt(2x^2))
where do i go from here to get real numbers or is there a better way?
15. I have the diagram labeled but i'm not sure how to find AB. should i use the right triangle ABD (if BD is the median to AC)?
thanks.
10 and 14. i don't even know where to start. any suggestions?
11. i started by finding the diagonal on the bottom face: sqrt(2x^2), if the sides of the cube=x and labeling the angle ABC theta.
tan(theta)=x/sqrt(2x^2)
--> theta=tan^-1(x/sqrt(2x^2))
where do i go from here to get real numbers or is there a better way?
15. I have the diagram labeled but i'm not sure how to find AB. should i use the right triangle ABD (if BD is the median to AC)?
thanks.
posted by Monica Moore at 5:44 PM
1 Comments:
For #11, if CA=x, then AB=sqrt(2), b/c AB is the diagonal of the base. angle ABC=tan^(-1)(x/sqrt(2x))=tan^(-1)(1/sqrt2).angle ABC=35.26degrees.
for #14, if the origin is O, then Ox=cos(theta), Oy=sin(theta). the base of triangle APC=2*Ox=2cos(theta), and the height=Oy+1=sin(theta+1). area=1/2*base*h=1/2*(2cos(theta))*(sin(theta)+1)=cos(theta)*(sin(theta)+1). then you can graph and find the maximum area.
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