CPhill

Using the information in the first part of the problem, let d be the distance from C to B and h be the height of the building. So we have,

tan(25) = h/d   Solving for h we have, h = d*tan(25)

And, since we have a right triangle with two legs d  and 160, the distnce that D is from B = √(160² + d²)

So we have

tan(20) = (h)/√(160² + d²) = (d*tan(25)) / √(160² + d²)

Simpifying, we have

tan(20)/ tan(25) = d/√(160² + d²)     and we can write

tan(25)/tan(20) = √(160² + d²) /d       Square both sides

[tan(25)/tan(20)]² = (160² + d²)/d²    and the right side = 160/d² + 1   Subtract 1 from both sides

([tan(25)/tan(20)]² - 1)  = 160²/d²        and we can write

d² = 160² / ([tan(25)/tan(20)]² - 1)   Simplifying this, we have

d² = 160² / (.6413959572703699169)     Now....take the square root of both sides

d = 199.7822378461935026188794268 m

And using  ....   h = d*tan(25) ....we have....

h = (199.7822378461935026188794268)* tan(25) ≈ 93.159 m or just  93 m

I think that's it.....I haven't worked one like this before....thanks for submitting it....I hope I haven't made any grevious mistakes !!  (P.S.   ....thanks to alan for catching my previous error.....I went back and corrected it so that now, it appears that I'm smarter than I might really be....also...rom's method is better and actually conforms to what was asked....but I guess my approach shows that there's more than one way to skin a cat !!)   (so to speak)

Actually...I believe the proper term is "oblique asymptote"....here's a page that might help:

I believe that the questioner might have also meant this (but I'm not sure).......  y = (x^2+6x-3)/(x-3)

heureka is correct.....there is no horizontal asymptote in a "high/low" rational function.......(I believe we have a "slant" asymptote, instead)

This says that we want to subtract f(x) from g(x) and put that result back into f(x).

So g(x) - f(x)  =    (3x + 2) - (2x - 1)    =   x + 3

So  f(x + 3)  =  2(x + 3) - 1  = 2x + 6 - 1  = 2x + 5

Check out the big brain on rosala.........!!!    thumbs up and points!!!

Triangles with equal sides are congruent. Then, since AO = CO and BO = DO and chord CD = chord AB, then triangle AOB = triangle COD. Thus, angle AOB = angle COD = 77 = x .

That one seems pretty easy, huh??

.....That's what I've been trying to TELL them!!

A isn't true....

To see why, note that triangle ACP is similar to triangle CBP. Therefore  angle CAP = angle BCP. And angle BCP = angle BCD and BCD subtends arc DB.  And angle CAP = angle CAB, and angle CAB subtends arc BC. But since CAP = BCP and  BCD = CAB, then arc DB = arc BC.  But arc BC is much greater than arc AC, therefore, arc BD is much greater than AC.

Just to add another proof to the mix....in any triangle, an angle opposite a greater side is larger than an angle opposite a lesser side. So, in triangle ACB, BC > AC. Therefore, angle CAB > CBA. Then, since they are bpth inscribed angles, the arc intercepted by CAB > the arc intercepted by CBA. And, as was shown above, the arc intercepted by CAB = the arc intercepted by BCD = BD. And the arc intercepted by CBA = AC. Thus, arc BD > arc AC.

Ha!!!  I can do some basic geometry stuff......just don't ask me any probability questions.....!!!

Also...I'm afraid I'm not a millionaire...but...if you wanted to send me a  check.......

Wow!!!...This one's a little tough, huh??  Let's see if we can figure it out !!!

The angle at the top of the circle is a right angle because it subtends an arc of 180. And an inscribed angle measures1/2 of its intercepted arc. So the measure  of "d" = 90 -53 = 36.

And since d measures 36 degrees, it's intercepted arc = 72. But this is the same arc subtended by the central angle "c."  And a central angle = the measure of its intercepted arc. So "c" = 72.

And the central angle "b" is just supplementary to "c" so it = 180 - 72 = 108.

And "a" subtends the same arc as "b" = 108. But since it's an inscribed angle, it's measure is (1/2) of this = 54.

I think that's it.......