This is a nice little brain teaser. I can't dedicate time to properly formatting my solutions right now, so I will have to do that later.
Oh, and I missed that there are 5 problems and not 4, so I will need to deal with that one, as well, but it should be easy enough, if my surface skimming of the text is not betraying me.
In problem 1, there are some terms that feel like they should cancel out, but they don't. I even went so far so as to do a sanity check using Wolfram.
In problem 2, the only tricky part is noticing that 0.008^(2*log_c(x)-1) is 0.2^(6*log_c(x)-3).
Problem 3 seems trivial, considering that log_c(x^0) = 0 if we are working with real logarithms, and 27 = 3^3.
Problem 4 just asks for your knowledge of the fact that tg2(x/2) = (1-cos2(x))/(1+cos2(x)). After you use the relevant identity, all you have left is to solve a rather simple equation with respect to cos(x).
Problem 5 - you need to basically just find the length of the lateral side of the trapezium. The angles of the trapezium by the larger base have the angular measure of (2*alpha+(Pi-2*alpha)/2)/2 = alpha/2+Pi/4, because of how angles on a circle depend on corresponding arcs. After that, finding the remaining angles is easy. From there, we can easily find the height of the trapezium, and from there we can find the area of the trapezium and the lateral side of it. From there, we multiply the length of the lateral side of the trapezium by tg(alpha) and get the height of the prism. After that, we multiply the area of the trapezium by the height of the prism and voila.
This is a nice little brain teaser. I can't dedicate time to properly formatting my solutions right now, so I will have to do that later.
Oh, and I missed that there are 5 problems and not 4, so I will need to deal with that one, as well, but it should be easy enough, if my surface skimming of the text is not betraying me.
Damn. I completely forgot about this.
However, none of the problems is hard.
In problem 1, there are some terms that feel like they should cancel out, but they don't. I even went so far so as to do a sanity check using Wolfram.
In problem 2, the only tricky part is noticing that 0.008^(2*log_c(x)-1) is 0.2^(6*log_c(x)-3).
Problem 3 seems trivial, considering that log_c(x^0) = 0 if we are working with real logarithms, and 27 = 3^3.
Problem 4 just asks for your knowledge of the fact that tg2(x/2) = (1-cos2(x))/(1+cos2(x)). After you use the relevant identity, all you have left is to solve a rather simple equation with respect to cos(x).
Problem 5 - you need to basically just find the length of the lateral side of the trapezium. The angles of the trapezium by the larger base have the angular measure of (2*alpha+(Pi-2*alpha)/2)/2 = alpha/2+Pi/4, because of how angles on a circle depend on corresponding arcs. After that, finding the remaining angles is easy. From there, we can easily find the height of the trapezium, and from there we can find the area of the trapezium and the lateral side of it. From there, we multiply the length of the lateral side of the trapezium by tg(alpha) and get the height of the prism. After that, we multiply the area of the trapezium by the height of the prism and voila.